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Problem solving with Computer
Algebraic Systems
Ákos Pintér
TÁMOP-4.1.2.D-12/1/KONV-2012-0008
Debrecen
2014
Preface
The purpose of this book is twofold. First we would like to provide a collection
from some harder elementary mathematical problems. Second we try to popularize
the Computer Algebraic Systems to solve exercises including IMO-level challenging
questions. Enjoy it!
1
Abbreviations
AIME – American Invitational Mathematics Examination
ARML – American Regional Mathematics League
CMO – Canadian Mathematical Olympiad
HMMT – Harvard-MIT Math Tournament
BW – Baltic Way
KöMaL – Középiskolai Matematikai Lapok
MOSP – Mathematical Olympiad Summer Program
USAMO – United States of America Mathematical Olympiad
IMOLL – International Mathematical Olympiad Long List
IMOSL – International Mathematical Olympiad Shortlist
JMOP – Japan Mathematical Olympiad Preliminary
TITEE – Tokyo Institute Of Technology Entrance Examination
FTST – France Team Selection Test
ARO – All-Russian Olympiad
IZO – International Zhautykov Olympiad
INMO – Iran National Math Olympiad
RNO – Romania National Olympiad
USAIMTS – USA International Mathematical Talent Search
SMT – Stanford Mathematics Tournament
PUMC – Princeton University Mathematical Competition
USAPC – USA Purple Comet
3
4
Chapter 1
Algebra and Number Theory
Problem (AIME, 1983/6) Let an equal 6n + 8n . Determine the remainder upon
dividing a83 by 49.
Solution: We will use the binomial theorem for expanding (7 − 1)83 and (7 + 1)83 .
We have
83
83 2
83 82
83
83
7−1
7 +
7 + ... −
(7 − 1) = 7 −
82
81
1
and
83
(7 + 1)
83
83 82
83 2
7 + 1.
7 +
=7 +
7 + ... +
82
81
1
83
Now for the sum
(7 − 1)83 + (7 + 1)83 ,
half of our terms cancel out and we are left with
2(783 + 3403 · 781 + · · · + 83 · 7).
One can realize that all of these terms are divisible by 49 except the final term
14 · 83. After some quick division, the answer is 35.
5
6
Solution with CAS: 683 + 883 mod49;
35
Problem (AIME, 1983/6) What is the largest 2-digit prime factor of the integer
200
?
100
Solution: Expanding the binomial coefficient, we get
200!
200
=
.
100
100!100!
Let the prime be p, then 10 ≤ p < 100. In the case p > 50, then the prime factor
p appears twice in the denominator. Thus, we need p to appear as a factor three
times in the numerator, so 3p < 200. The largest such prime is 61.
);
Solution with CAS: ifactor( 200
100
23 · 3 · 5 · 11 · 132 · 17 · 37 · 53 · 59 · 61 · 101 · · · 199
Problem (AIME, 1986/8) Let S be the sum of the base 10 logarithms of all the
proper divisors (all divisors of a number excluding itself) of 1000000. What is the
integer nearest to S?
Solution: The factorization of 1000000 = 26 56 , so it has
(6 + 1)(6 + 1) = 49
divisors, of which 48 are proper. The sum of multiple logarithms of the same base
is equal to the logarithm of the products of the numbers. Writing out the first few
terms, we see that the answer is equal to
log 1 + log 2 + log 4 + . . . + log 1000000 = log(20 50 )(21 50 )(20 51 ) · · · (26 56 ).
Each power of 2 appears 7 times and the same goes for 5. So the overall power of
7
2 and 5 is
7(1 + 2 + 3 + 4 + 5 + 6) = 7 · 21 = 147.
Ssince the question asks for proper divisorsonly, we have to exclude 26 56 , so each
power is actually 141 times. The answer is thus
S = log 2141 5141 = log 10141 = 141.
Solution with CAS: A := divisors(1000000); P := 1; for i to nops(A) do;
P := P ∗ A[i]; end do; print(log[10](P ));
147
It means that the sum of the base 10 logarithms of all the proper divisors is
147 − 6 = 141.
Problem (AIME, 1987/3) By a proper divisor of a natural number we mean a
positive integral divisor other than 1 and the number itself. A natural number
greater than 1 will be called nice if it is equal to the product of its distinct proper
divisors. What is the sum of the first ten nice numbers?
Solution: Let p(n) denote the product of the distinct proper divisors of n. It is
easy to see that a number n is nice if and only if
(i) it has exactly two distinct prime divisors.
(ii) it is the cube of a prime number.
Indeed, suppose that another nice number existed that does not fall into one
of these two categories. Then we can either express it in the form n = pqr (with
p, q prime and r > 1) or n = pe (with e 6= 3). In the former case, it suffices to note
that
p(n) ≥ pr · qr = pqr2 > pqr = n.
In the latter case, then
p(n) = 1 · p · p2 · · · pe = pe(e+1)/2
8
. For p(n) = n, we have
pe(e+1)/2 = pe
that is e2 + e = 2e. This gives us e = 0, 3 (the case e = 0 does not work). Thus,
listing out the first ten numbers to fit this form,
2 · 3 = 6, 23 = 8, 2 · 5 = 10, 2 · 7 = 14, 3 · 5 = 15, 3 · 7 = 21,
2 · 11 = 22, 2 · 13 = 26, 33 = 27, 3 · 11 = 33.
The sum of these numbers is 182.
Solution with CAS: We print the nice numbers under 50.
B := {}; for i from 2 to 100 do; A:= divisors(i);
k:=nops(A); j:= product(A[l], l = 2..k − 1);
if i = j then B:= ‘union‘(B, {i}); end if; end do; print(B);
{6, 8, 10, 14, 15, 21, 22, 26, 27, 33, 34, 35, 38, 39, 46}
Problem (AIME, 1986/5) Find 3x2 y 2 if x and y are integers such that
y 2 + 3x2 y 2 = 30x2 + 517.
Solution: If we move the 30x2 term to the left side, we can factorize:
(3x2 + 1)(y 2 − 10) = 517 − 10
The prime factorization of 507 is equal to 3 · 132 . Since x and y are integers,
3x2 + 1 clearly cannot equal a multiple of three. 169 does not work either, so we
obtain 3x2 + 1 = 13, and x = ±2. This yields y 2 − 10 = 39, so y = ±7. The asked
product
3x2 y 2 = 3 × 4 × 49 = 588.
9
Solution with CAS: factor(3x2 y 2 + y 2 − 30x2 − 10);
(y 2 − 10)(3x2 + 1)
Problem (AIME, 1986/5) Compute
(104 + 324)(224 + 324)(344 + 324)(464 + 324)(584 + 324)
.
(44 + 324)(164 + 324)(284 + 324)(404 + 324)(524 + 324)
Solution: We will apply the identity
x4 + 4y 4 = (x2 + 2y 2 − 2xy)(x2 + 2y 2 + 2xy)
for x4 + 324. Now we have
x4 + 324 = (x2 − 6x + 18)(x2 + 6x + 18)
and thus
104 + 324 = 58 · 178, 224 + 324 = 370 · 634,
344 + 324 = 970 · 1378, 464 + 324 = 1858 · 2410, 584 + 324 = 3034 · 3730,
further,
44 + 324 = 10 · 58, 164 + 324 = 178 · 370,
284 + 324 = 634 · 970, 404 + 324 = 1378 · 1858, 524 + 324 = 2410 · 3034.
Finally we get that the original fraction is 373.
Solution with CAS: (104 + 324)(224 + 324)(344 + 324)(464 + 324)(584 + 324)/((44 +
324)(164 + 324)(284 + 324)(404 + 324)(524 + 324));
373
Problem (AIME, 1989/3) Suppose n is a positive integer and d is a single digit
10
in base 10. Find n if
n
= 0.d25d25d25 . . . .
810
Solution: We sum an infinite geometric series,
∞
X
100d + 25
d25
=
.
0.d25d25d25 . . . =
n
1000
999
i=1
Thus we have
100d + 25
n
=
810
999
so
100d + 25
4d + 1
= 750
.
37
37
Since 750 and 37 are relatively prime, 4d + 1 must be divisible by 37, and the only
digit for which this is possible is d = 9. Thus 4d + 1 = 37 and n = 750.
n = 30
Solution with CAS: for d from 1 to 9 do; if type((1/999) ∗ (810 ∗ (100 ∗ d +
25)), integer) = true then print(d); end if; end do;
9
Problem (AIME, 1996/2) For each real number x, let ⌊x⌋ denote the greatest
integer that does not exceed x. For how many positive integers n is it true that
n < 1000 and that
⌊log2 n⌋
is a positive even integer?
Solution: For integers k, we have to consider the equation
⌊log2 n⌋ = 2k,
or
2k ≤ log2 n < 2k + 1
11
This implies
22k ≤ n < 22k+1 .
Thus, n must satisfy these inequalities (since n < 1000):
4 ≤ n < 8,
16 ≤ n < 32,
64 ≤ n < 128,
and
256 ≤ n < 512.
There are 4, 16, 64, and 256 solution of the first, second, third and fourth inequality,
respectively, so the answer is 4 + 16 + 64 + 256 = 340.
Solution with CAS: s := 0; for n from 1 to 999 do;
if type((1/2)*floor(log[2](n)), integer) then s := s + 1;
end if; end do; print(s);
341
Problem (AIME, 1998/1) For how many values of k is 1212 the least common
multiple of the positive integers 66 , 88 , and k?
Solution: It is trivial that k has only 2 and 3 in its prime factorization, i. e.
k = 2a 3b . The least common multiple of any numbers an be found by writing
out their factorizations and taking the greatest power for each factor. One can
calculate
[66 , 88 ] = 224 36 .
Therefore
1212 = 224 · 312 = [224 36 , 2a 3b ] = 2max(24,a) 3max(6,b) ,
and b = 12. Since 0 ≤ a ≤ 24, there are 25 values of k.
Solution with CAS: s := 0; for a from 0 to 24 do; for b from 0 to 12 do;
12
k := 2a ∗ 3b ; if lcm(66 , 88 , k) = 1212 then s:= s+1; end if; end do; end
do; print(s);
25
Problem (AIME, 2002, I/4) Consider the sequence defined by
ak =
k2
1
+k
for k ≥ 1. Given that
am + am+1 + · · · + an−1 =
1
,
29
for positive integers m and n with m < n, find m + n.
Solution: One can calculate that
k2
1
1
1
1
=
= −
.
+k
k(k + 1)
k k+1
Thus we obtain
am + am+1 + · · · + an−1 =
1
1
1
1
1
1 1
1
−
+
−
+· · ·+
− = − .
m m+1 m+1 m+2
n−1 n
m n
This yields
n−m
1
= .
mn
29
Easy to see that 29|mn, and without loss of generality we may assume that n = 29t
for some integer t. Sustituting,
29t − m = mt,
29t = (t + 1)m,
29t
= m,
t+1
13
and finally,
29
= m.
t+1
Since m is an integer, t + 1 = 29, t = 28. It quickly follows that n = 28 · 29 and
m = 28, so m + n = 840.
29 −
Solution with CAS: for n from 1 to 10000 do;
if type(1/(1/29 + 1/n), integer) = true then print(n);
end if; end do;
Problem (AIME, 2002, II/6) Find the integer that is closest to
1000
10000
X
n=3
n2
1
.
−4
Solution: We will use the formula
n2
So if you pull the
1
4
4
1
1
=
−
.
−4
n−2 n+2
out of the summation, you get
250
10,000 X
n=3
1
1
−
n−2 n+2
.
Now that telescopes, leaving you with:
1 1 1
1
1
1
1
250 1 + + + −
−
−
−
2 3 4 9997 9998 9999 10000
=
1
1
1
1
−
−
−
= 250 + 125 + 83.3 + 62.5 − 250 −
9997 9998 9999 10000
Now,
1
1
1
1
−
−
−
250 −
9997 9998 9999 10000
is not enough to bring 520.8 lower than 520.5 so the answer is 521.
.
14
Solution with CAS: evalf(1000(sum(1/(k 2 − 4), k = 3..10000)));
520.7333383
Problem (AIME, 2002, II/7) It is known that, for all positive integers k,
12 + 22 + 32 + . . . + k 2 =
k(k + 1)(2k + 1)
.
6
Find the smallest positive integer k such that
12 + 22 + 32 + . . . + k 2
is a multiple of 200.
Solution: It is clear that
k(k + 1)(2k + 1)
6
is a multiple of 200 if k(k + 1)(2k + 1) is a multiple of
1200 = 24 · 3 · 52 .
So 16, 3, 25 are divisors of k(k +1)(2k +1). Since 2k +1 is always odd, and only one
of k and k + 1 is even, either k, k + 1 ≡ 0 (mod 16). Thus, k ≡ 0, 15 (mod 16).
If k ≡ 0 (mod 3), then 3|k. If k ≡ 1 (mod 3), then 3|2k +1. If k ≡ 2 (mod 3),
then 3|k + 1. Thus, there are no restrictions on k in (mod 3). One can see that
only one of k, k + 1, and 2k + 1 is divisible by 5. So either k, k + 1, 2k + 1 ≡ 0
(mod 25). Thus, k ≡ 0, 24, 12 (mod 25). From the Chinese Remainder Theorem,
we have
k ≡ 0, 112, 224, 175, 287, 399 (mod 400).
Thus, the smallest positive integer k is 112.
Solution with CAS: a :=sum(n2 , n = 1..k); for k to 1000 do; if a (mod 200) =
15
0 then print(k); end if; end do;
112, 175, 224 . . .
Problem (AIME, 2003, I/1) Given that
((3!)!)!
= k · n!,
3!
where k and n are positive integers and n is as large as possible, find k + n.
Solution: A straightforward calculation shows that
((3!)!)!
(6!)!
720!
720!
720 · 719!
=
=
=
=
= 120 · 719!
3!
3!
3!
6
6
We certainly can not make n any larger if k is going to stay an integer, so the
answer is k + n = 120 + 719 = 839.
Solution with CAS: First we compute the smallest positive integer n such that
a = ((3!)!)!
is smaller than n!.
3!
for n to 1000 do if a/factorial(n) < 1 then print(n); end if; end do;
One can see that the smallest positive integer n with the above mentioned property
a
, and we obtain 120.
is 720. Now we determine the fraction 719!
Problem (AIME, 2003, I/9) An integer between 1000 and 9999, inclusive, is called
balanced if the sum of its two leftmost digits equals the sum of its two rightmost
digits. How many balanced integers are there?
Solution: If the common sum of the first two and last two digits is n, 1 ≤ n ≤ 9,
there are n choices for the first two digits and n + 1 choices for the second two
digits (since zero may not be the first digit). This gives
9
X
n=1
n(n + 1) = 330
16
balanced numbers. If the common sum of the first two and last two digits is
n, 10 ≤ n ≤ 18, there are 19 − n choices for both pairs. This gives
18
X
n=10
2
(19 − n) =
9
X
n2 = 285
n=1
balanced numbers. Thus, there are in total 330 + 285 = 615 balanced numbers.
Solution with CAS: s := 0; for a to 9 do; for b from 0 to 9 do; for c
from 0 to 9 do; for d from 0 to 9 do; if a + b = c + d then s := s + 1;
end if; end do; end do; end do; end do;print(s);
615
Problem (AIME, 2005, II/5) Determine the number of ordered pairs (a, b) of
integers such that
loga b + 6 logb a = 5,
with 2 ≤ a ≤ 2005, and 2 ≤ b ≤ 2005.
Solution: Our equation can be rewritten as
log a
(log b)2 + 6(log a)2
log b
+6·
=
= 5.
log a
log b
log a log b
On multiplying through by log a log b and factoring yields
(log b − 3 log a)(log b − 2 log a) = 0.
Therefore, log b = 3 log a or log b = 2 log a, so either b = a3 or b = a2 . In the first
case b = a2 , note that 442 = 1936 and 452 = 2025. Thus, all values of a from
2 to 44 will work. For the case b = a3 , note that 123 = 1728 while 133 = 2197.
Therefore, for this case, all values of a from 2 to 12 work. There are 44−2+1 = 43
possibilities for the square case and 12 − 2 + 1 = 11 possibilities for the cubic case.
Thus, the answer is 43 + 11 = 54.
17
Solution with CAS: s := 0; for a from 2 to 2005 do; for b from 2 to 2005
do; if loga (b)+6logb (a) = 5 then s := s+1; end if; end do; end do;print(s);
54
Problem (AIME, 2005, II/7) Let
4
√
√
√
.
x= √
4
( 5 + 1)( 5 + 1)( 8 5 + 1)( 16 5 + 1)
Find (x + 1)48 .
Solution: We will use the following formula
(
√
2n
5 + 1)(
√
2n
5 − 1) = (
√
2n
5)2 − 12 =
√
2n−1
5 − 1.
It now becomes apparent that if we multiply the numerator and denominator of
4
√
√
√
√
4
( 5 + 1)( 5 + 1)( 8 5 + 1)( 16 5 + 1)
√
√
by ( 16 5 − 1), the denominator will telescope to 1 5 − 1 = 4, so
√
√
4( 16 5 − 1)
16
x=
= 5 − 1.
4
It follows that
(x + 1)48 = (
√ 48
5) = 53 = 125
16
.
Solution with CAS: x:= (4(root[16](5.) − 1))/
((sqrt(5.) + 1)(root[4](5.) + 1)(root[8](5.) + 1)(root[16](5.) + 1)(root[16](5.) −
1));
x := 0.1058230170
18
(x + 1)48 ;
124.9999998
Problem (AIME, 2006 II/3) Let P be the product of the first 100 positive odd
integers. Find the largest integer k such that P is divisible by 3k .
Solution: The product of the first 100 positive odd integers can be written as
1 · 3 · 5 · 7 · · · 195 · 197 · 199 =
1 · 2 · · · 200
200!
= 100
.
2 · 4 · · · 200
2 · 100!
Hence, we seek the number of threes in 200! decreased by the number of threes
in 100!. Using the Legendre formula on the prime factorizations of a factorial we
have
200
200
200
+
+
+ . . . = 97
3
9
27
threes in 200! and
100
100
100
+
+
+ . . . = 48
3
9
27
threes in 100! Therefore, we have a total of 97 − 48 = 49 threes.
Solution with CAS: s := 1; for k from 3 to 199 by 2 do; s := s · k; end do;
ifactor(s);
349 · 525 · · ·
Problem (AIME, 2006 II/3) There exist unique positive integers x and y that
satisfy the equation
x2 + 84x + 2008 = y 2 .
Find x + y.
Solution: Completing the square, we have
y 2 = x2 + 84x + 2008 = (x + 42)2 + 244.
19
Thus we obtain
244 = y 2 − (x + 42)2 = (y − x − 42)(y + x + 42)
by difference of squares. Since 244 is even, one of the factors is even. A parity check
shows that if one of them is even, then both must be even. Since 244 = 22 · 61,
the factors must be 2 and 122. Since x, y > 0, we have y − x − 42 = 2 and
y + x + 42 = 122. Finally, the latter equation implies that x + y =80. Indeed, by
solving, we find (x,y) = (18,62) is the unique solution.
Solution with CAS: ifactor(327);
{3}{109}
Problem (AIME, 2009 II/7) Define n!! to be
n(n − 2)(n − 4) · · · 3 · 1
for n odd and
n(n − 2)(n − 4) · · · 4 · 2
for n even. When
2009
X
(2i − 1)!!
i=1
(2i)!!
is expressed as a fraction in lowest terms, its denominator is 2a b with b odd. Find
ab
.
10
Solution: First, we note that
(2n)!! = 2n · n!,
and
(2n)!! · (2n − 1)!! = (2n)!.
20
We can now take the fraction
(2i − 1)!!
(2i)!!
and multiply both the numerator and the denominator by (2i)!!. We get that this
fraction is equal to
(2i)!
(2i)!
= 2i 2 .
2
(2i)!!
2 (i!)
2i
, hence this fraction is
is
simply
Now we can observe that (2i)!
2
(i!)
i
2i
i
22i
and our sum turns into
S=
2009
X
i=1
Let
c=
2009 X
2i
i=1
i
,
2i
i
22i
.
· 22·2009−2i .
c
. Hence if S is expressed
Obviously c is an integer, and S can be written as 22·2009
as a fraction in lowest terms, its denominator will be of the form 2a for some
a ≤ 2 · 2009. In other words, we just showed that b = 1. To determine a, we need
to determine the largest power of 2 that divides c. Let p(i) be the largest x such
that 2x that divides i. We can now return to the observation that
(2i)! = (2i)!! · (2i − 1)!! = 2i · i! · (2i − 1)!!.
Together with the obvious fact that (2i − 1)!! is odd, we get that
p((2i)!) = p(i!) + i.
It immediately follows that
2i
p
= p((2i)!) − 2p(i!) = i − p(i!),
i
21
and hence
2i
2·2009−2i
= 2 · 2009 − i − p(i!).
·2
p
i
Obviously, for i ∈ {1, 2, . . . , 2009} the function
f (i) = 2 · 2009 − i − p(i!)
is is a strictly decreasing function. Therefore
p(c) = p
2 · 2009
= 2009 − p(2009!).
2009
We can now compute
p(2009!) =
∞ X
2009
k=1
2k
= 1004 + 502 + · · · + 3 + 1 = 2001.
Hence p(c) = 2009 − 2001 = 8. And thus we obtain
a = 2 · 2009 − p(c) = 4010,
and the answer is
4010 · 1
ab
=
= 401.
10
10
Solution with CAS: a :=sum(((2i − 1)!!)/((2i)!!), i = 1..2009);
6693254562 . . . 13981551
134984099 . . . 812161024
ifactor(denom(a));
24010
22
Problem (AIME, 2010 I/2) Find the remainder when
is divided by 1000.
9 × 99 × 999 × · · · × |99 {z
· · · 9}
999 9’s
Solution: One can see that
999 ≡ 9999 ≡ . . . ≡ 99
· · · 9} ≡ −1
| {z
999 9’s
(mod 1000).
That is a total of 999 − 3 + 1 = 997 integers, so all those integers multiplied out
are congruent to −1 (mod 1000). Thus, the entire expression is congruent to
(−1)(9)(99) = −891 ≡ 109
(mod 1000).
Solution with CAS: mul(10i − 1, i = 1..999) mod 1000;
109
Problem (AIME, 2010 II/6) Find the smallest positive integer n with the property
that the polynomial
x4 − nx + 63
can be written as a product of two non constant polynomials with integer coefficients.
Solution: We have to distinguish two ways for a monic fourth degree polynomial to
be factored into two non-constant polynomials with real coefficients: into a cubic
and a linear equation, or 2 quadratics.
Consider the case when the factors are cubic and linear. Let x − r1 be the
linear factor, where r1 is a zero of the given quartic, and let
x3 + ax2 + bx + c
23
be the cubic. By the rational roots theorem, we have r1 = 1, 3, 7, 9, or 63. One
can calculate that
(x3 + ax2 + bx + c)(x − r1 ) = x4 + (a − r1 )x3 + (b − ar1 )x2 + (c − br1 )x − cr1 .
Noe setting the appropriate coefficients equal, we obtain a−r1 = 0, thus a = r1 ,
further, by b − ar1 = 0, b = a2 . Finally, c − br1 = −n implies n = a3 − c, and
−cr1 = 63 We get c = −63
. It follows that
a
n = a3 +
63
,
a
where a = r1 = 1, 3, 7, 9, or 63, which reach minimum when a = r1 = 3, where
n = 48.
In the second case the factors are quadratics. Let
x2 + ax + b
and
x2 + cx + d
be the two quadratics, so that
(x2 + ax + b)(x2 + cx + d) = x4 + (a + c)x3 + (b + d + ac)x2 + (ad + bc)x + bd.
Therefore, again setting coefficients equal, a + c = 0, a = −c, b + d + ac =
0, b + d = a2 , and ad + bc = −n, finally, bd = 63. Since b + d = a2 , the only possible
values for (b, d) are (1, 63) and (7, 9). From this we find that the possible values
for n are ±8 · 62 and ±4 · 2. Therefore, the answer 8.
Solution with CAS: for n from 1 to 10000 do; if irreduc(x4 − n ∗ x + 63) =
false then print(n): end if: end do;
8
24
48
64
352
496
736
9264
Problem (AIME, 2012 II/1) Find the number of ordered pairs of positive integer
solutions (m, n) to the equation
20m + 12n = 2012.
Solution: Solving for m gives us
m=
503 − 3n
,
5
so in order for m to be an integer, we must have
3n ≡ 503
(mod 5)
which implies
n≡1
(mod 5).
The smallest possible value of n is obviously 1, and the greatest is
503 − 5
= 166,
3
so the total number of solutions is
166−1
5
+ 1 = 34.
Solution with CAS: s := 0; for m from 1 to 101 do; for n from 1 to 167
25
do; if 20∗m+12∗n = 2012 then s := s+1; end if; end do; end do;print(s);
34
Problem (KöMaL Gy.3142.) What is the smallest positive integer divisible by
28, which ends 28 (in base 10) and the sum of its digits is 28?
Solution: Because the last two digits of the appropriate number is 28 it has the
form 100k + 28. The number is divisible by 28, and thus by 7. We have that 100k
and so k is also divisible by 7. The sum of the digits of 100k is 18, so 100k (and
k) is divisible by 9. We have k is divisible by 7 and 9, that is by 63. We consider
the first few numbers with these properties:
6328, 12628, 18928, . . . .
One can see that the smallest number with our conditions is 18928.
Solution with CAS: for a to 9 do; for b from 0 to 9 do; for c from 0 to
9 do;
k :=a ∗ 104 + b ∗ 103 + c ∗ 102 + 28;
if ‘and‘(‘mod‘(k, 28) = 0, a+b+c = 18) then print(k); end if; end do;
end do; end do;
18928
37828
56728
69328
75628
81928
88228
26
94528
Problem (KöMaL B.3353.) Determine
i
h √
2 1 · 2 · 3 + 2 · 3 · 4 + . . . + 1998 · 1999 · 2000 .
Solution: First we consider the number under square root:
1 · 2 · 3 + 2 · 3 · 4 + . . . + 1998 · 1999 · 2000 =
1999
1999
1999
X
X
X
(i − 1)i(i + 1) =
i3 −
i.
i=2
i=2
i=2
Using the identities
k
X
i3 =
i=1
and
k
X
i=1
we have
i=
k 2 (k + 1)2
,
4
k(k + 1)
,
2
1 · 2 · 3 + 2 · 3 · 4 + . . . + 1998 · 1999 · 2000 =
19992 · 20002 1999 · 2000
−
,
4
2
and
4(1 · 2 · 3 + 2 · 3 · 4 + . . . + 1998 · 1999 · 2000) = 19992 · 20002 − 2 · 1999 · 2000.
Finally we obtain that the required value is 1999 · 2000 − 1 = 3997999.
Solution with CAS: floor(2· sqrt(sum (i · (i + 1) · (i + 2), i = 1..1998)))
= 3997999
27
Remark: A classical result by Bernoulli is the formula
1k + 2k + . . . + (x − 1)k =
1
(Bk+1 (x) − Bk+1 (0)) ,
k+1
where Bk+1 (X) denotes the k + 1th Bernoulli polynomial with degree k + 1.
Problem (KöMaL K.58.) How many perfect square divisors are of the number
857304000?
Solution: The canonical form of 857304000 is
857304000 = 26 · 37 · 53 · 72 .
An arbitrary divisor of this number has a form
d = 2α · 3β · 5γ · 7δ ,
where
0 ≤ α ≤ 6, 0 ≤ β ≤ 7, 0 ≤ γ ≤ 3, 0 ≤ δ ≤ 2.
A divisor is square if and only if every exponent is even in its canonical form. So
we get 4, 4, 2, 2 possibilities for α, β, γ and δ, respectively, that is we have 64
perfect square divisors.
Solution with CAS: with(numtheory);
A:=divisors(857304000);
k := 0; for i to nops(A) do; if type(sqrt(A[i]), integer) = true then
k := k+1;
end if; end do; print(k);
64
Problem (KöMaL B.4196.) Let n be a positive integer. Determine the first digit
28
after decimal point of the number
n
X
k(k + 1)
n
k=1
.
Solution: One can prove (by mathematical induction) that
12 + 22 + . . . + n2 =
n(n + 1)(2n + 1
,
6
and
1 + 2 + ... + n =
n(n + 1)
.
2
Using these formulas we have
n
X
k(k + 1)
k=1
n
=
(n + 1)(n + 2)
.
3
Now, it is easy to see that if n is not divisible by 3, then our sum is an integer
number, and if n is divisible by 3 then the first digit after decimal point is 6.
Solution with CAS: factor(sum(k*(k+1), k = 1 ..
n))
1
n(n + 1)(n + 2)
3
Problem (KöMaL B.4378.) Let p be a positive prime number. Solve the equation
x3 y 3 + x3 y 2 + x2 y 3 + x2 y 2 − x + y = p + 2
in the set of integer numbers.
Solution: The idea of our solution is based on the formula
x3 y 3 + x3 y 2 − x2 y 3 + x2 y 2 − x + y − 2 = p
29
and on the factorization of the polynomial on the left hand side. It is easy to check
that
x3 y 3 + x3 y 2 − x2 y 3 + x2 y 2 − x + y − 2 = (xy − 1)(x2 y 2 + x2 y − xy 2 + 2xy + x − y + 2),
so we get xy−1 = ±1 or xy−1 = ±p, and the other factor is ±p or ±1, respectively.
First case: xy − 1 = 1. Then xy = 2 and we have
x2 y 2 + x2 y − xy 2 + 2xy + x − y + 2 = 4 + 2(x − y) + 4 + x − y + 2 = 3(x − y) + 10.
In the case x = 1, y = 2 we obtain 3(x − y) + 10 = 7, and if x = 2, y = 1, then
3(x − y) + 10 = 13, further, if x = −1, y = −2, then 3(x − y) + 10 = 13, and finally
if x = −2, y = −1, then 3(x − y) + 10 = 7, and we get the pairs
(x, y) = (1, 2), (2, 1), (−1, −2), (−2, −1)
as the solutions of our problems.
Second case: xy − 1 = −1. Now we get xy = 0, that is x = 0 or y = 0. In the first
case −y + 2 = −p, that is y = p + 2, in the second case x + 2 = −p, which implies
x = −p − 2. The corresponding solutions are
(x, y) = (0, p + 2), (−p − 2, 0).
Third case: xy − 1 = p. Then the factor
x2 y 2 + x2 y − xy 2 + 2xy + x − y + 2
is 1. However,
x2 y 2 + x2 y − xy 2 + 2xy + x − y + 2 = (xy + 1)(xy + 1 + x − y) + 1,
which gives
(xy + 1)(xy + 1 + x − y) = 0
30
Since xy = p + 1, we have p + 2 + x − y = 0. Substituting this into the previous
equation gives
x(p + 2 + x) = p + 1
which is impossible.
Finally, if xy − 1 = −p, then
(xy + 1)(xy + 1 + x − y) + 1 = −1,
which shows that
(2 − p)(1 − p + x − y) = −2
and thus p = 3 and (x, y) = (1, −2), or (2, −1).
Solution with CAS: factor(x3 ∗ y 3 + x3 ∗ y 2 − x2 ∗ y 3 + x2 ∗ y 2 − x + y − 2);
(y ∗ x − 1) ∗ (x2 ∗ y 2 + y ∗ x2 − y 2 ∗ x + 2 ∗ y ∗ x + x − y + 2)
Problem (KöMaL K.311.) One can see that the integer
772009 + 772010 + 772011 + 772012
is divisible by 7 and 11. Determine all the prime divisors of this number.
Solution: It is easy to see that
772009 + 772010 + 772011 + 772012 = 772009 (1 + 77 + 772 + 773 ) = 462540 · 772009 ,
and 462540 is divisible by 3, 4, and 5. A short calculation gives the prime factors
13 and 593. Now, all the prime divisors of our numbers are
{2, 3, 5, 7, 11, 13, 593}.
31
Solution with CAS: ifactor (772009 + 772010 + 772011 + 772012 );
22 · 3 · 5 · 72009 · 112009 · 13 · 593.
Problem (KöMaL K.18.) Determine the distinct digits a, b, c and d such that
abcd : dcba = 4,
where abcd and dcba denote four-digit numbers.
Solution: From the condition we have
4 · dcba = abcd.
Rewriting this for digits we obtain
1000a + 100b + 10c + d = 4000d + 400c + 40b + 4a
and
332a + 20b = 1333d + 130c.
One can see that d = 1 or d = 2 and (since d is even) we have d = 2. The number
2a − 3d = 2a − 6 is divisible by 10-zel, a − 3 is is divisible by 5, so a = 3 or 8,
however, a is at least 4, finally we get a = 8. These values yield
2656 + 20b = 2666 + 130c,
that is
2b = 1 + 13c.
The unique solution of this equation is (b, c) = (7, 1).
Solution with CAS: for a to 9 do; for b from 0 to 9 do; for c from 0 to
9 do; for d from 1 to 9 do; if 1000 ∗ a + 100 ∗ b + 10 ∗ c + d = 4 ∗ (1000 ∗
32
d + 100 ∗ c + 10 ∗ b + a) then print(1000 ∗ a + 100 ∗ b + 10 ∗ c + d) end if; end
do; end do; end do; end do;
Problem (KöMaL C.781.) Determine the positive prime numbers p > q > r, for
which
p2 − (q + r)2 = 136.
Solution: We can factorize the left hand side
(p + q + r)(p − q − r) = 136 = 23 · 17.
Suppose that r > 2, then the factor p+q +r is odd, and thus we have p+q +r = 17
and p − q − r = 8, which is obviously impossible. The contradiction gives r = 2.
Then we obtain two possibilities
p + q + 2 = 34, p − q − 2 = 4,
or
p + q + 2 = 68, p − q − 2 = 8.
In the first case we get p = 19, q = 13, and the second case yields p = 38, q = 28.
Solution with CAS: with(numtheory);
divisors(136);
{1, 2, 4, 8, 17, 34, 68, 136}
Problem (KöMaL C.854.) Prove that the following equation
13 + 33 + 53 + . . . + (2n − 1)3
= 2n2 − 1
1 + 3 + 5 + . . . + (2n − 1)
is true for every positive integer n.
33
Solution: One can prove by mathematical induction that the formulas
13 + 23 + . . . + n3 =
n2 (n + 1)2
,
4
and
n(n + 1)
.
2
are true for every positive integer n. Using these identities we get
1 + 2 + ... + n =
(2n)2 (2n + 1)2
n2 (n + 1)2
1 + 3 + 5 + . . . + (2n − 1) =
−8·
,
4
4
3
3
3
3
and
1 + 3 + 5 + . . . + (2n − 1) =
further,
2n(2n + 1)
n(n + 1)
−2·
,
2
2
n2 (2n + 1)2 − 2n2 (n + 1)2
n2 (2n2 − 1)
=
.
n(2n + 1) − n(n + 1)
n2
Solution with CAS:
sum((2k − 1)3 , k = 1..n)
factor
sum(2k − 1, k = 1..n)
= 2n2 − 1.
Problem (KöMaL B.3671.) Solve the equation
(x2 + y)(x + y 2 ) = (x − y)3
in the set of integer numbers.
Solution: If y = 0, then every integer x is a solution of our equation. In the sequel
we assume that y 6= 0. Reordering the equation and dividing it by y we have the
quadratic (in y) equation
2y 2 + (x2 − 3x)y + (3x2 + x) = 0
Since y is integer, the discriminant of this quadratic equation is a perfect square,
34
i. e.
(x2 − 3x)2 − 4 · 2 · (3x2 + x) = x(x − 8)(x + 1)2
is a perfect square. This expression is a perfect square if and only if x(x − 8) is a
perfect square so we have to solve the equation
x(x − 8) = k 2
in unknown integers x, k. From this equation we get
(x − 4)2 − k 2 = (x + k − 4)(x − k − 4) = 16
and (x, k) = (−1, ±3), (0, 0), (8, 0), (9, ±3).
Solution with CAS: factor((x2 + y)(y 2 + x) − (x − y)3 );
y(yx2 + 2y 2 + x + 3x2 − 3yx)
solve(yx2 + 2y 2 + x + 3x2 − 3yx = 0, y);
3
1√ 4
3
1√ 4
1
1
x − 6x3 − 15x2 − 8x, − x2 + x −
x − 6x3 − 15x2 − 8x
− x2 + x +
4
4
4
4
4
4
factor(x4 − 6x3 − 15x2 − 8x);
x(x − 8)(x + 1)2
Problem (KöMaL C.1087.) The first term of an arithmetic progression is 1, the
second term is n, further, the sum of the first n terms is 33n. Determine n.
Solution: The difference of our arithmetic progression is n − 1. Using the formula
for the sum of the first n terms of an arbitrary arithmetic progression we have
Sn =
n(a1 + an )
n(1 + 1 + (n − 1)2 )
=
= 33n,
2
2
35
which gives n = 9.
Solution with CAS: solve((n − 1)2 + 2 = 66, n);
9, −7
Problem (KöMaL K.13.) Determine the last two digits of the sum
71 + 72 + . . . + 72005 .
Solution: Applying the well-known formula for the sum of first n terms of a geometric progression we obtain
71 + 72 + . . . + 72005 =
72006 − 7
.
6
The binomial theorem gives
2006
7
1003
= (50 − 1)
1003
2500 + 1003 · 50 − 1 = 1000B + 649,
= 1000A −
2
that is the last two digits of the number
72006 − 7
6
are 42 : 6 = 07.
Solution with CAS: sum (7i , i = 1..2005);
30797610223128525661 . . . 2262083769819607
Problem (KöMaL C.778.) For an arithmetic progression Sm denotes the sum of
36
the first m terms. Prove that
Sn+k
Sn − Sk
=
n+k
n−k
for every n > k ≥ 1.
Solution: It is known that
Sn+k =
(n + k)(a1 + an+k )
(n + k)(2a1 + (n + k − 1)d)
=
,
2
2
and similarly
Sn =
n(2a1 + (n − 1)d)
k(2a1 + (k − 1)d)
, Sk =
,
2
2
From these relations we have
Sn − Sk =
2a1 (n − k) + (n(n − 1) − k(k − 1)) d
.
2
However,
n(n − 1) − k(k − 1) = (n − k)(n − k − 1),
which proves our statement.
Solution with CAS: f := (2a1 + (n + k − 1)d)(n + k)/(2(n + k)) − ((2a1 + (n −
1)d)n − (2a1 + (k − 1)d)k)/(2(n − k)); simplify(f );
0
Problem (KöMaL C.752.) Let a, b, c be consecutive terms of a geometrical progression. Prove that the numbers
a + b + c,
p
3(ab + bc + ca),
√
3
27abc
are also consecutive terms of another geometrical progression.
37
Solution: Since positive numbers a, b, c are consecutive terms of a geometrical
progression we have ac = b2 . Now we have to show that
√
3
27abc · (a + b + c) = 3(ab + bc + ac).
We know that
27abc = 27b3
and
√
3
27abc = 3b,
so the right hand side is
3b(a + b + c) = 3(ab + bc + b2 ) = 3(ab + bc + ac).
Solution with CAS: factor(3b(a + b + c) − 3(ab + bc + ca));
3b2 − 3ca
Problem (KöMaL C.984.) Three, not necessarily consecutive terms of an arithmetic progression containing only positive numbers are a, b and c. We know that
c−b a−c b−a
+
+
= 0.
a
b
c
Determine the difference of the sequence!
Solution: On multiplying through we have
0 = bc(c − b) + ac(a − c) + ab(b − a) = bc2 − b2 c + a2 c − ac2 + ab(b − a) =
c2 (b − a) − c(b2 − a2 ) + ab(b − a) = (b − a)(c2 − c(a + b) + ab) =
(b − a)(c − a)(c − b).
This yields a = b or a = c or b = c so the difference of our arithmetic sequence is
38
0.
Solution with CAS: factor(bc(c − b) + ac(a − c) + ab(b − a));
−(−c + b)(a − c)(−b + a)
Problem (KöMaL C.1064.) The real numbers x, y, z are consecutive terms of a
non-constant arithmetic progression such that
cos x + cos y + cos z = 1,
and
1
sin x + sin y + sin z = √ .
2
Determine the tangent of the twelfth term if the first term is x.
Solution: Squaring the equations we obtain
cos2 x + cos2 y + cos2 z + 2 cos x cos y + 2 cos x cos z + 2 cos y cos z = 1
and
1
sin2 x + sin2 y + sin2 z + 2 sin x sin y + 2 sin x sin z + 2 sin y sin z = ,
2
then taking their sum and using the formulas
sin2 α + cos2 = 1,
and
cos α cos β + sin α sin β = cos(α − β),
we have
3
3 + 2 cos(x − y) + 2 cos(x − z) + 2 cos(y − z) = .
2
Since the numbers x, y, z are consecutive terms of a non-constant arithmetic prog-
39
ression, thus
y = x + d, z = x + 2d,
with d 6= 0. Finally, we get
3 + 2 cos(−d) + 2 cos(−2d) + 2 cos(−d) =
which gives
3
2
3
2 cos d + cos 2d = − .
4
It is known that
cos 2α = 2 cos2 α − 1,
so
1
= 0.
4
There is only one solution of the quadratic equation with the condition cos d ≥ −1
2 cos2 d + 2 cos d −
cos d =
r
3 1
− .
8 2
Using x = y − d and z = y + d in the first equation of our problem, we have
cos(y − d) + cos d + cos(y + d) = 1,
and
r
2
1
cos y =
=
.
1 + 2 cos d
3
From these equations one can calculate the values d and y, and the twelfth term
is y + 10d.
Solution with CAS: solve(2x2 + 2x − 1/4 = 0, x);
1 1√
1 1√
− +
6, − −
6
2 4
2 4
Problem (KöMaL B.3529.) The sum of the first few terms of a geometrical
40
progression is 11, the sum of their squares is 341, and the sum of their cubes is
3641. Determine the terms of geometrical progression.
Solution: From the conditions we have
a1 + a2 + . . . + an = 11,
a21 + a22 + . . . + a2n = 341,
and
a31 + a32 + . . . + a3n = 3641.
Using the formula for the sum of first n terms of an arbitrary geometrical progression we obtain
qn − 1
a1
= 11,
q−1
a21
and
a31
We remark that q 6= 1, otherwise
q 2n − 1
= 341,
q2 − 1
q 3n − 1
= 3641.
q3 − 1
an = 11, a2 n = 341,
. Now we divide the the second and the third equation by the
which yields n = 11
31
first equation we obtain
qn + 1
a1
= 31,
q+1
and
a21
q 2n + q n + 1
= 331.
q2 + q + 1
Since
a1 (q n − 1) = 11(q − 1)
41
and
a1 (q n + 1) = 31(q + 1),
an easy calculation gives
a1 = 10q + 21,
and
qn =
11(q − 1)
21q + 10
+1=
.
10q + 21
10q + 21
(It is clear that a1 6= 0). Substituting these values into the equation
a21
q 2n + q n + 1
= 331
q2 + q + 1
we arrive at
2q 2 + 5q + 2 = 0,
so q1 = −2, q2 = − 12 and
a1 = 1, n = 5
or
a1 = 16, n = 5,
respectively.
Solution with CAS: solve(2q 2 + 5q + 2 = 0, q);
1
− , −2
2
Problem (KöMaL C.701.) Prove that the number
1 · 2 · . . . · 1001 + 1002 · 1003 · . . . · 2002
is divisible by 2003.
Solution: The remainder of 1002 dividing by 2003 is equal to the remainder of
42
1002 − 2003 = −1001 dividing by 2003. The remainder of 1003 dividing by 2003
is equal to the remainder of 1003 − 2003 = −1000 dividing by 2003, and so on. So
the remainder of the number
1 · 2 · . . . · 1001 + 1002 · 1003 · . . . · 2002
is equal to the remainder of the number
1 · 2 · . . . · 1001 + (−1001)(−1000)(−999) · . . . · (−1)
dividing by 2003. One can see that this number is 0, so our statement is proved.
Solution with CAS: (product(i, i = 1 .. 1001)+product(i, i = 1002 .. 2002), 2003)
mod 2003
0
Problem (KöMaL C.704.) Determine the positive integers n for which
log2 3 · log3 4 · . . . · logn (n + 1) = 10?
Solution: It is known that
loga b =
1
.
logb a
Using this formula we have
log2 3 · log3 4 =
log3 4
= log2 4,
log3 2
log2 3 · log3 4 · log4 5 = log2 4 · log4 5 =
log4 5
= log2 5,
log4 2
and finally
log2 3 · log3 4 · . . . · logn (n + 1) = log2 n · logn (n + 1) =
logn (n + 1)
= log2 (n + 1).
logn 2
43
Now we have to solve the equation
log2 (n + 1) = 10
which gives n = 1023 as a solution.
Solution with CAS: evalf(product(log[i](i + 1), i = 2..1023));
10
Problem (KöMaL B.3542.) Prove that a number in the form 1111 . . . 11 is divisible by 7, then is is also divisible by 37.
Solution: From the condition we know that
7|1111 . . . 11 =
10n − 1
.
9
It shows that the exponent n is divisible by 6, that is the number is divisible by
106 − 1. However,
106 − 1 = 33 · 7 · 11 · 13 · 37,
so our statement is proved.
Solution with CAS: We calculate the exponent n between 1 and 40 such that the
n
(mod ) 7 =
number 10 9−1 is divisible by 7. for i to 40 do; if (10i − 1)/9
0 then print(i) end if; end do; The result: {6, 12, 18, 24, 30, 36}. From this
we can conjecture that the numbers consisting 1’s are divisible by 7 if and only if
the corresponding exponent is divisible by 6. Indeed, in this case the number is
divisible by 37.
Problem (KöMaL B.3522.) Solve the equation
2x4 + x2 y 2 + 5y 2 = y 4 + 10x2
44
in the set of integer numbers.
Solution: Reordering the equation we have
2x4 + x2 y 2 + 5y 2 − y 4 − 10x2 .
One can observe that
2x4 + x2 y 2 + 5y 2 − y 4 − 10x2 = (2x2 − y 2 )(x2 + y 2 − 5) = 0.
From this 2x2 = y 2 , and x = y = 0 or x2 + y 2 = 5. The last equation possesses 8
integer solutions
x = ±2, y = ±1, x = ±1, y = ±2,
so the original equation has 9 solutions in the set of integer numbers.
Solution with CAS: factor(2x4 + x2 y 2 + 5y 2 − y 4 − 10x2 );
(2x2 − y 2 )(x2 + y 2 − 5)
Problem (KöMaL B.3530.) Calculate the exact value of the expression
1001
2000
2001
2002
.
− ... −
+
−
1001
2
1
0
Solution: Let n be a positive integer and let Sn denote the sum
[n/2]
n−i
.
(−1)
i
i=0
X
i
We have to calculate S2002 . One can prove, by using the identity
n
n−1
n−1
=
+
,
k
k−1
k
45
that
Sn = Sn−1 + Sn−2
for every positive integer n ≥ 3. A straightforward calculation gives that
S1 = S6 = 1, S2 = S5 = 0,
and
S3 = S4 = −1.
By mathematical induction we have that
S6k = S6k+1 = 1, S6k+2 = S6k+5 = 0, S6k+3 = S6k+4 = −1,
for every positive integer k. Since 2002 = 6 · 333 + 4, as a final result we obtain
S2002 = −1.
Solution with CAS: sum((−1)i binomial(2002 − i, i), i = 0..1001);
−1
Problem (KöMaL B.3517.) Prove that 71|61! + 1.
Solution: The Wilson’s Theorem gives that 70! + 1 is divisible by 71. To prove our
statement is enough to prove that the number
61! − 70!
is divisible by 71. It is clear that
61! − 70! = 61!(1 − 62 · 63 · 64 . . . · 69 · 70),
and an easy calculation shows
1 − (−9)(−8) · . . . · (−1) = 9! + 1 = 19 · 71 · 269.
46
Solution with CAS: ifactor (61! + 1);
71 · 227 · 48795702665964504883 · 645414773564183 . . . 58905144895791
Problem (KöMaL C.648.) What is the exact value of the expression
2log6 18 · 3log6 3 ?
Solution: Using some well-known identities of the logarithm function we have
log6 18 = log6 3 + log6 6 = log6 3 + 1,
and thus
2log6 18 · 3log6 3 = 2 · 2log6 3 · 3log6 3 = 2 · 6log6 3 = 6.
Solution with CAS: evalf(2log[6](18) ∗ 3log[6](3) );
6.000000001
Problem (KöMaL K.252.) We multiply the sum of 6 consecutive integer numbers
by the sum of the next 6 consecutive numbers. Prove that the remainder of the
product dividing by 36 is independent on the choice of the original integer numbers.
Solution: Let
6k, 6k + 1, . . . , 6k + 5,
and
6k + 6, 6k + 7, . . . , 6k + 11
denote the integers. The corresponding sums are 36k+15 and 36k+51, respectively,
and the product is
(36k + 15)(36k + 51) = 362 k 2 + 66 · 36k + 15 · 51.
47
This form yields that the remainder dividing by 36 is 9 for every k.
Solution with CAS: ‘mod‘((36k + 15)(36k + 51), 36);
9
Problem (KöMaL C.595.) Determine all the three-digit numbers such that they
are equal to the sum of factorials of their digits.
Solution: Let N = abc denote an integer with the above mentioned property. Then
we have
100a + 10b + c = a! + b! + c!.
It is clear that the number N does not contain the digits 7, 8, 9, because the
factorial of these digits are greater than 1000. If the number N contains the digit
6 then it contains at least a digit 7, which is impossible. The number N must
contain a digit 5, because 3 · 4! = 72 < 100. If it contains only one digit 5 then
the number is between 5! + 2 · 0! and 5! + 2 · 4!, so the number begins with digit 1,
further, the factorial of the third digit is 1, 2, 6 or 24. The possible candidates are
121 + 1, 121 + 2, 121 + 6, 121 + 24,
and a quick check shows that the unique solution is 145.
If the number contains two digits 5 then the number is between 240 + 0! and
240 + 4! that is it begins with digit 2. The sum of the factorials is 242, however
this number does not contain any digit 5.
Finally, if N contains three digits 5, we have N = 555, however,
3 · 5! 6= 555.
There is only one solution of the problem, namely, 145 = 1 + 4! + 5!.
Remark: There is a straightforward generalization of our problem. Determine the
positive integers such that they are equal to the sum of factorials of their digits.
48
Let k be an n-digit number with this property. Then
k ≥ 10n ,
on the other hand the sum of factorials of digits is at most
n · 9! = 362800n.
Thus we have
10n ≤ 363800n.
and this gives n ≤ 6.
Solution with CAS: for a to 9 do; for b from 0 to 9 do; for c from 0 to
9 do; if 100∗a+10∗b+c = factorial(a)+factorial(b)+factorial(c) then
print(a, b, c); end if; end do; end do; end do;
1, 4, 5
Problem (KöMaL C.570.) Determine the three-digit prime numbers such that
the product of their digits is 189.
Solution: The canonical form of 189 is 33 · 7. Since we have digits, the prime
numbers with the property mentioned above consist of the digits 3, 7, 9. We have
six possibilities, from these numbers
379, 397, 739, 937
are prime numbers and 793 = 13 · 61 and 973 = 7 · 139 are composite numbers.
Solution with CAS: for a to 9 do; for b from 0 to 9 do; for c from 0 to
9 do; n := 100a + 10b + c; if (isprime(n) = true) and (abc = 189) then
print(n); end if; end do; end do; end do;
49
Problem (KöMaL C.573.) Solve the following equation
12 + 22 + . . . + n2 = 1 + 2 + . . . + (2n − 1) + 2n
in the set of positive integers.
Solution: It is known that
12 + 22 + . . . + n2 =
n(n + 1)(2n + 1)
,
6
and
1 + 2 + . . . + (2n − 1) + 2n =
2n(2n + 1
.
2
From these formulas we obtain
n(n + 1)(2n + 1)
= n(2n + 1),
6
and n = 5.
Solution with CAS: factor(sum(k 2 , k = 1..n)-(sum(k, k = 1..2n)));
1
n(2n + 1)(n − 5)
6
Problem (KöMaL B.3342.) Determine the positive integers n such that the expression n2 − 19n + 99 is perfect square.
Solution: We have to solve the diophantine equation
n2 − 19n + 99 = m2
On multiplying by 4, we have
4n2 − 76n + 396 = (2n − 18)2 − 324 + 396 = (2n − 18)2 + 72 = (2m)2 .
50
and after reordering and factorization we get
(2m − 2n + 18)(2m + 2n − 18) = 72,
and
(m − n + 9)(m + n − 9) = 18.
An easy calculation gives n = 1, 9, 10 and 18.
Solution with CAS: for n from 1 to 30 do; if type(sqrt(n2 − 19 ∗ n + 99),
integer) = true then print(n); end if; end do;
1
9
10
18
Problem (HMMT, 2005) We know that the number 27000001 has four distinct
prime divisors. What is their sum?
Solution: Here we use the formula
a3 + b3 = (a + b)(a2 + ab + b2 ).
It is clear that
27000001 = 3003 + 13 = 301 · (3002 − 300 + 1) =
= 7 · 43 · (3012 − 302 ) = 7 · 43 · 271 · 331.
Solution with CAS: ifactor(27000001);
{7, 43, 271, 331}
51
Problem (ARML, 2002) Let a be an integer number such that
1+
1 1
1
a
+ + ... +
=
.
2 3
23
23!
What is the remainder of a dividing by 13?
Solution: The equation yields
1 1
1
a = 23! 1 + + + . . . +
2 3
23
.
. The
It is clear that every summand is divisible by 13, apart from the term 23!
13
remainder of this number dividing by 13 is equal to the remainder of the number
1 · 2 · · · · 6 · (−6) · (−5) · (−1) · 1 · 2 · · · · 10
dividing by 13. This is the remainder of
(6!)3 · 7 · 8 · 9 · 10
dividing by 13. Since 6! = 720 = 55 · 13 + 5, the remainder is 7.
Solution with CAS: ‘mod‘(factorial(23)*(sum(1/n, n = 1..23)), 13)
7
Problem (MOSP, 1998) Prove that the sum of 3, 4, 5 or 6 consecutive square is
never square. Give an example that the sum of 11 consecutive squares is square.
Solution: Suppose that the sum of three consecutive squares is square. This means
that the equation
3n2 + 2 = (n − 1)2 + n2 + (n + 1)2 = y 2
52
has a solution in unknown integers n and y. It is known that the remainder of a
square dividing by 3 is 0 or 1. The left hand side gives 2 as a remainder dividing
by 3, so there is no solution of the equation.
We consider the sum of four consecutive squares, that is consider the equation
(n − 1)2 + n2 + (n + 1)2 + (n + 2)2 = 4n2 + 4n + 6 = y 2
in n and y unknown integers. Since the left hand side is even, and thus y 2 and y
are even. In this case 4 is a divisor of y 2 , however, the 2-order of the left hand side
is 1. We arrive at a contradiction, there is no solution.
If we consider the sum of 5 consecutive squares then we get the equation
(n − 2)2 + (n − 1)2 + n2 + (n + 1)2 + (n + 2)2 = 5n2 + 10 = y 2
The left hand side is divisible by 5, so 5|y, and 25|y 2 = 5(n2 +2). This yields 5|n2 +
2, however, a square give 0, 1 or 4 as a remainder dividing by 5, a contradiction.
Considering the sum of 6 consecutive squares yields the equation
(n − 2)2 + (n − 1)2 + n2 + (n + 1)2 + (n + 2)2 + (n + 3)2 = 6n2 + 6n + 19 = y 2 .
It is known that a square gives 0 or 1 as a remainder dividing by 4. The left hand
side is
6n2 + 6n + 19 = 6n(n + 1) + 19.
Since n or n + 1 is even, the left hand side give 3 as a remainder dividing by 4, a
contradiction again.
Finally, the sum of 11 consecutive squares leads to the equation
11n2 + 2(12 + 22 + 32 + 42 + 52 ) = 11n2 + 110 = y 2 .
One can see that n = 1, y = 11 is a solution of this equation. Indeed,
(−4)2 + (−3)2 + (−2)2 + (−1)2 + 02 + 12 + 22 + 32 + 42 + 52 + 62 = 121 = 112 .
53
Solution with CAS: for n from 1 to 1000 do: if type(sqrt(11 ∗ n2 + 110),
integer) = true then print(n): end if; end do;
1
23
43
461
859
6
4
Problem (KöMaL B.4449.) How many zeros does the number 45 + 65 (in base
10) end with?
4
Solution: Since the number is divisible by 25 , so we consider its 5-order. The
binomial theorem shows
6
56
4
56
= (5 − 1)
=
5 6
X
5
i
i=0
and
6 −i
5i (−1)5
4
54
6
54
= (5 + 1)
=
5 4
X
5
i=0
i
5i .
It is clear that for i = 0, in the first sum is −1 van, in the second sum is 1, so their
sum is 0. If i = 1, the the corresponding sum is 56 + 55 = 6 · 55 . We will prove
that for i > 1, the summands
6
5
6
5i (−1)5 −i
i
and
4
5
5i
i
54
are divisible by 56 . This statement is trivial for i ≥ 6. If 2 ≤ i ≤ 4, then we have
k
5k (5k − 1) · · · (5k − i + 1)
5
=
,
i!
i
that is this number is divisible by 5k , and thus
6
5
6
5i (−1)5 −i
i
and
4
5
5i
i
is divisible by 56 . Finally, in the case i = 5, one can see that the binomial
coefficients
6 4
5
5
,
i
i
are divisible by 5. This yields that the sum is divisible by 55 , but it is not divisible
by 56 , that is the number ends with 5 zeros.
6
4
Solution with CAS: 45 + 65 ;
1539446141412626239132 . . . 484800000
Problem (KöMal C.1160.) What is the remainder of
20122013 + 20132012
dividing by 2012 · 2013?
Solution: Consider the fraction
20122013 + 20132012
.
2012 · 2013
55
Dividing by terms we have
20122013 + 20132012
20122012 20132011
=
+
.
2012 · 2013
2013
2012
Now we apply the binomial theorem and we get
20122012 = (2013 − 1)2 012 = 2013A + (−1)2012 = 2013A + 1,
where A is a positive integer and similarly, we have
20132011 = (2012 + 1)2011 = 2012B + 12011 ,
where B is a positive integer. Using this notation
2013A + 1 2012B + 1
20122013 + 20132012
=
+
=
2012 · 2013
2013
2012
1
1
2013 + 2012
+B+
=A+B+
,
2013
2012
2012 · 2013
that is the remainder is 4025.
A+
Solution with CAS: 20122013 + 20132012 mod 2012 · 2013;
4025
Remark: A straightforward generalization of our problem is the following. What
is the remainder of the sum
nn+1 + (n + 1)n
dividing by n(n + 1)? Consider the fraction
nn+1 + (n + 1)n
.
n(n + 1)
56
Dividing by terms we have
nn+1 + (n + 1)n
nn
(n + 1)n−1
=
+
.
n(n + 1)
n+1
n
Since
nn = (n + 1 − 1)n = A(n + 1) + (−1)n
and
(n + 1)n = Bn + 1,
where A and B are positive integers, we obtain
(n + 1)n−1
(−1)n
1
nn
+
=A+
+B+ ,
n+1
n
n+1
n
so the remainder is
(−1)n n + n + 1.
Problem (KöMal B.4503.) Determine the four-digit perfect square numbers such
that their first and last two digits are equal.
Solution: Our number is four-digit one and its first two and last two digits are
equal, we have to find it in the form aabb, and
x2 = aabb
for some integer x. Since x2 is a four-digit number we have
1000 ≤ x2 ≤ 9999,
thus
21 ≤ x ≤ 99.
It is clear that
x2 = 1000a + 100a + 10b + b = 11(100a + b),
57
so we have 11|x2 , and 11|x. We obtain 7 possibilities for x
x ∈ {33, 44, 55, 66, 77, 88, 99}
and a quick calculation gives the unique solution 882 = 7744.
Solution with CAS: for a to 9 do; for b from 0 to 9 do; n := a(103 +102 )+
11b; if type(sqrt(n), integer) = true then print(n); end if; end do;
end do;
88
Problem (KöMal K.356.) Two digits of the six-digit number
2X01X2
are missing. Find the missing digits such that the corresponding six-digit number
is divisible by 36 and 117.
Solution: Since 36 = 4 · 9 and 117 = 9 · 13, our six-digit integer is divisible by 4, 9
and 13. The number is divisible by 4, so the the last two digits in our number is
divisible by 4, and thus the next to last digit (nld) is 1, 3, 5, 7 or 9 lehet. Since the
six-digit number is divisible by 9, the sum of its digits is also divisible by 9. Now,
if nld is 1, then the second digit is 3, if nld is 3, then the second digit is 1, if nld
is 5, then the second digit is 8, if nld is 7, then the second digit is 6, and, finally,
if nld 9, then the second digit is 4. The corresponding numbers are
230112, 210132, 280152, 260172, 240192.
We know that the required number is divisible by 13. After a short calculation we
get the unique solution 210132, so the missing digits are 1 and 3.
Solution with CAS: for a from 0 to 9 do for b from 0 to 9 do; n := 2 ·
105 + a · 104 + 100 + 10b + 2; if n mod 36 · 13 = 0 then print(n); end if;
end do; end do;
58
Problem (KöMal Gy.3254.) Is there a perfect cube in the form
ababab1
(in base 10)?
Solution: Let n = k 3 be an integer number in the form ababab1. Since
106 ≤ n < 107 ,
so we have
100 ≤ k ≤ 215.
Expanding the form of this number we get
n = 1 + b · 10 + a · 100 + b · 1000 + a · 104 + b · 105 + a · 106 =
= 1 + b(10 + 103 + 105 ) + a(102 + 104 + 106 ).
Now we have
n = 1 + (10 + 103 + 105 )(b + 10a),
and thus n − 1 is divisible by 30. Since
n − 1 = k 3 − 1 = (k − 1)(k 2 + k + 1)
is divisible by 30, we will show that k − 1 is also divisible by 30. Because the cube
of this number is odd, we have that k is odd and so k − 1 is even. The cube of
k is in the form 3l + 1, where l is an integer number, so k is in the form 3l + 1.
Indeed, it is clear that k is not divisible by 3, and if it would be in the form 3l + 2,
the its cube also would be in the form 3l + 2. If k is divisible by 5, then k 3 − 1 is
not divisible by 5, further if k = 5l + 2, then k 3 − 1 = (5l + 2)3 − 1 gives 2 as a
remainder dividing by 5, if k = 5l + 3, then k 3 − 1 gives 1 as a remainder dividing
5, and finally, if k = 5l + 4, then k 3 − 1 yields 3 as a remainder dividing 5. We
59
have k = 5l + 1 and from this we have k = 30l + 1. In the interval
100 ≤ k ≤ 215
we get for possibilities for k, namely,
k ∈ {121, 151, 181, 211}.
A final check gives the unique solution
2113 = 9393931.
Solution with CAS: for a from 1 to 9 do; for b from 0 to 9 do n := a ·
106 + b · 105 + a · 104 + b · 103 + a · 102 + 10b + 1; if type(root[3](n), integer)
= true then print(n); end if; end do; end do;
Problem (KöMaL B.4243.) Prove that the number
6564 + 64
is composite.
Solution: Completing the square gives
6564 + 64 = (6532 + 8)2 − 2 · 8 · 6532 = (6532 + 8)2 − (4 · 6516 )2 .
Applying the identity
a2 − b2 = (a − b)(a + b)
and we get
6564 + 64 = (6532 − 4 · 6516 + 8)(6532 + 4 · 6516 ),
that is our number is composite.
60
Solution with CAS: ifactor(6564 + 64) =
3916733 · 12653×
×848064026205894987841230736413208938269543326969×
×2500297 · 89009 · 136573 · 929501×
×89509365027181240985382554457790961
Problem (KöMaL C.1001.) Determine the positive integer such that it has two
prime divisors, the number of its divisors is 6, and the sum of its divisors is 28.
Solution: We consider the numbers in the form
n = pα q β ,
where p és q are distinct prime numbers,
(α + 1)(β + 1) = 6,
further σ(n) = 28, where σ(n) denotes the sum of positive divisors of n. It is easy
to see that n has the form pq 2 , so
1 + p + pq + pq 2 + q + q 2 = 28.
It is clear that q ≤ 3, otherwise the sum above is larger than 28. If q = 2, then
1 + p + 2p + 4p + 2 + 4 = 28,
and this yields p = 3. If q = 3, then
1 + p + 3p + 9p + 3 + 9 = 28,
és from this we have 13p = 15, which is impossible. The solution of our problem
61
is n = 12.
Solution wit CAS: with(numtheory); for i from 1 to 27 do; A := divisors(i);
k := nops(A); B := 0; for j from 1 to k do B := B+A[j]; end do; if B
= 28 then print(i, B); end if; end do;
12, 28
62
Chapter 2
Equations and Inequalities
Problem (KöMaL B.4433.) Solve the equation (1 + x)8 + (1 + x2 )4 = 82x4 .
Solution: Since x = 0 is not a solution of our equation we can divide by x4 . We
obtain
4 4
1
1
x+ +2 + x+
= 82.
x
x
We introduce a new variable, let y = x + x1 + 1. Rewriting the previous equation
we have
(y + 1)4 + (y − 1)4 = 82,
and expanding it we have the following biquadratic equation
2(y 4 + 6y 2 − 40) = 0
This gives that y 2 = −10 or y 2 = 4. It is clear that the first value is impossible
and the second value yields
1
x + = −3
x
or
1
x + = 1.
x
However, the second case is impossible, because for an arbitrary nonzero real
63
64
number x we have
1
1
≥ 2 orx + ≤ −2,
x
x
depending on the sign of x. So the real solutions of our original equations coincide
the solutions of the quadratic equation x2 + 3x + 1 = 0.
x+
4
Solution with CAS: factor (x + 1)8 + (1 + x2 ) − 82 x4 ;
2 x2 + 3 x + 1 x2 − x + 1 x4 + 2 x3 + 13 x2 + 2 x + 1
Since x2 − x + 1 > 0 and
x4 + 2 x3 + 13 x2 + 2 x + 1 = x2 (x + 1)2 + 11x2 + (x + 1)2 > 0
for every real x, we obtain the above mentioned result.
Problem (KöMaL B.4027.) Solve the equation
x2 + 1
1
=
2
x + 11
6
r
11x − 6
.
6−x
Solution: Taking the square of both sides and expanding the expressions we get
the equation
47x5 − 222x4 + 314x3 − 564x2 + 1367x − 942 = 0.
It is easy to see that x = 1, 2 and x = 3 are solutions of the equation, so we can
factorize the polynomial of degree 5. Indeed,
47x5 − 222x4 + 314x3 + 1367x − 942 = (x − 1)(x − 2)(x − 3)(47x2 + 60x + 157).
Now, the discriminant of the quadratic polynomial is negative, that is all potential
real solutions of the equation are 1, 2 és 3. Please check them!
65
Figure 2.1: Graph of the function
x2 +1
x2 +11
−
1
6
q
11x−6
6−x
Solution with CAS: factor(47x5 − 222x4 + 314x3 − 564x2 + 1367x − 942);
(x − 1)(x − 2)(x − 3)(47x2 + 60x + 157)
Remark: One can give a graphical solution, see Figure 2.1.
Problem (KöMaL C.1118.) Solve the equation
4x2 +
in the set of real numbers.
√
3
=2 x
4
66
Solution: Squaring and reordering the original equation we have
256x4 + 96x2 − 64x + 9 = 0.
One can see that x =
1
4
is a solution of the original equation, so we get
256x4 + 96x2 − 64x + 9 = (4x − 1)(64x3 + 16x2 + 28x − 9)
further,
1
4
is a zero of the polynomial of degree 3, we can factorize
64x3 + 16x2 + 28x − 9 = (4x − 1)(16x2 + 8x + 9).
The quadratic polynomial is positive for every real x, so we have only one solution
x = 14 .
Solution with CAS: See Figure 2.2!
Problem (KöMaL B.3403.) Prove that
1
1 1 1
= + +
a+b+c
a b c
implies
a5
1
1
1
1
= 5 + 5 + 5.
5
5
+b +c
a
b
c
for every a, b, c.
Solution: Using the condition
1 1 1
1
= + + ,
a+b+c
a b c
we get that the product
(a + b)(a + c)(b + c)
is 0. It means that a = −b or b = −c or a = −c which guarantees our statement.
67
Figure 2.2: The graphs of 4x2 +
3
4
√
and 2 x
68
1
a+b+c
Solution with CAS: factor
−
further,
factor
1
a5 +b5 +c5
−
1
a5
−
1
b5
1
a
− 1b −
1
c
=
(c + b)(c + a)(a + b)
,
(a + b + c)abc
−
−
−
1
c5
;
(c + b)(c + a)(a + b) · · ·
(a5 + b5 + c5 )a5 b5 c5
Remark: One can generalize our problem: Suppose that for real a, b, c
1
1 1 1
= + + ,
a+b+c
a b c
then prove that
1
a2k+1
+
b2k+1
+
c2k+1
=
1
a2k+1
+
1
b2k+1
+
1
c2k+1
,
where k is a positive integer.
Problem (KöMaL C.601.) Solve the equation
x2 + x +
√
x2 + x + 7 = 5
in the set of real numbers.
√
Solution: We will introduce a new variable, let y = x2 + x + 7 (The expression
under square root is positive for every x). Rewriting our equation we have
y 2 − 7 + y = 5,
that is
y 2 + y − 12 = 0
which gives y1 = −4, y2 = 3. The negative value is impossible, from the positive
69
Figure 2.3: The graph of x2 + x +
√
x2 + x + 7 − 5
value gives
x2 + x − 2 = 0
and we get x1 = 1, x2 = −2.
Solution with CAS: solve ((x2 + x − 5)2 − x2 − x − 7 = 0) ;
1 1√
1 1√
37, − +
37
−2, 1, − −
2 2
2 2
See Figure 2.3!
Problem (KöMaL B.4057.) Solve the equation
x6 − 6x + 5 = 0.
70
Solution: It is easy to see that x = 1 is a solution of our equation. Thus we can
factorize the polynomial as
x6 − 6x + 5 = (x − 1)(x5 + x4 + x3 + x2 + x − 5).
Observe that x = 1 is a zero of polynomial x5 + x4 + x3 + x2 + x − 5, so we obtain
x6 − 6x + 5 = (x − 1)2 (x4 + 2x3 + 3x2 + 4x + 5).
We show that
x4 + 2x3 + 3x2 + 4x + 5 > 0
for every real x. Indeed,
x4 + 2x3 + 3x2 + 4x + 5 = x2 (x + 1)2 + 2(x + 1)2 + 3 ≥ 3.
Solution with CAS: factor(x6 − 6x + 5);
(x − 1)2 (x4 + 2x3 + 3x2 + 4x + 5)
and solve(x4 + 2.x3 + 3x2 + 4x + 5);
{.287... + 1.416...I, −1.287... + .857I, −1.2878 − .857I, .287 − 1.416I}
The polynomial has four non-real zeros, and this fact gives that x4 + 2x3 + 3x2 +
4x + 5 > 0 for every real x.
Remark: See Figure 2.4!
Problem (AIME, 1983/3) What is the product of the real roots of the equation
√
x2 + 18x + 30 = 2 x2 + 18x + 45?
Solution: We substitute a new variable y, say, for x2 + 18x + 30. Our equation
71
Figure 2.4: The graph of x6 − 6x + 5
72
becomes
p
y = 2 y + 15.
Now we can square and solving for y, we have y = 10 or y = −6. The second
√
solution is extraneous (since 2 y + 15 is always positive). Thus we get y = 10 as
the unique solution for y. On substituting
x2 + 18x + 30
back in for y, we obtain
x2 + 18x + 30 = 10
and
x2 + 18x + 20 = 0.
Using the Vieta’s formulas, the product of the roots is 20.
Solution with CAS: solve(x2 + 18x + 30 − 2sqrt(x2 + 18x + 45) = 0, x);
−9 −
√
61, −9 +
√
61
Problem (AIME, 1983/5) Suppose that the sum of the squares of two complex
numbers x and y is 7 and the sum of the cubes is 10. What is the largest real
value that x + y can have?
Solution: We have
x2 + y 2 = (x + y)2 − 2xy = 7
and
x3 + y 3 = (x + y)(x2 − xy + y 2 ) = (7 − xy)(x + y) = 10.
Let w = x + y and z = xy. Using these substitutions we get
w2 − 2z = 7
73
and
w(7 − z) = 10.
We have to calculate the largest possible (real) value for w, thus we have to find
2
an expression for z in terms of w. w2 − 7 = 2z implies z = w 2−7 , and substituting,
we obtain
w3 − 21w + 20 = 0.
One can factorize the polynomial on the left hand side using the Rational Root
Theorem and
w3 − 21w + 20 = (w − 1)(w + 5)(w − 4) = 0,
and the largest possible solution is therefore x + y = w = 4.
Solution with CAS: factor(x3 − 21x + 20);
(x − 1)(x + 5)(x − 4)
Problem (AIME, 1983/9) Find the minimum value of
9x2 sin2 x + 4
x sin x
for 0 < x < π.
Solution: Let y = x sin x. Using this substitution we can rewrite the original
expression as
9y 2 + 4
4
= 9y + .
y
y
Since x > 0 and sin x > 0 because 0 < x < π, we have y > 0. We can apply the
arithmetical-geometrical mean, and we have
r
4
4
9y + ≥ 2 9y · = 12
y
y
74
Figure 2.5: The graph of
9x2 sin2 x+4
x sin x
The equality holds when
9y =
4
4
2
⇐⇒ y 2 = ⇐⇒ y = .
y
9
3
Therefore, the asked minimum value is 12 (when x sin x = 23 , since the function
x sin x is continuous and increasing on the interval 0 ≤ x ≤ π2 and its range on
that interval is from 0 ≤ x sin x ≤ π2 , by the Intermediate Value Theorem, this
value is attainable).
Solution with CAS: See Figure 2.5!
Problem (AIME, 1984/13) Find the value of
10 cot(cot−1 3 + cot−1 7 + cot−1 13 + cot−1 21).
75
Solution: We will use the fact
tan(arctan(x)) = x
and the addition formula,
tan(x + y) =
tan(x) + tan(y)
.
1 − tan(x) tan(y)
Now, let
a = cot−1 (3), b = cot−1 (7), c = cot−1 (13), d = cot−1 (21).
We obtain
1
tan(a) = ,
3
thus
1
tan(b) = ,
7
tan(c) =
1
,
13
tan(d) =
1
,
21
+ 17
1
tan(a + b) =
1 =
2
1 − 21
1
3
and
tan(c + d) =
so
1
13
1
+ 21
1
1 = ,
8
1 − 273
+ 18
2
tan((a + b) + (c + d)) =
1 = .
3
1 − 16
1
2
Finally, we get
10 cot(cot−1 3 + cot−1 7 + cot−1 13 + cot−1 21) = 10 ·
3
= 15.
2
Solution with CAS: expand(10ctg(arcctg3 + arcctg7 + arcctg13 + arcctg21));
15
76
Problem (AIME, 1986/1) What is the sum of the solutions to the equation
√
4
x=
12
√ ?
7− 4x
Solution: Introduce a new variable, let y =
√
4
x. Then we have
y(7 − y) = 12
and, by simplifying,
y 2 − 7y + 12 = (y − 3)(y − 4) = 0.
√
This means that 4 x = y = 3 or 4 and the asked sum of the possible solutions for
x is 44 + 34 = 337.
Solution with CAS: solve(root[4](x) = 12/(7-root[4](x)), x)
81, 256
Problem (AIME, 1986/4) Determine 3x4 + 2x5 if x1 , x2 , x3 , x4 , and x5 satisfy the
system of equations below.
2x1 + x2 + x3 + x4 + x5 = 6
x1 + 2x2 + x3 + x4 + x5 = 12
x1 + x2 + 2x3 + x4 + x5 = 24
x1 + x2 + x3 + 2x4 + x5 = 48
x1 + x2 + x3 + x4 + 2x5 = 96
Solution: Adding all five equations, we have
6(x1 + x2 + x3 + x4 + x5 ) = 6(1 + 2 + 4 + 8 + 16)
77
so
x1 + x2 + x3 + x4 + x5 = 31.
On subtracting this from the fourth and fifth given equation we obtain x4 = 17
and x5 = 65, respectively.
3x4 + 2x5 = 3 · 17 + 2 · 65 = 181.
Solution with CAS: solve(x[1] + x[2] + x[3] + x[4] + 2x[5] = 96, x[1] + x[2] + x[3] +
2x[4] + x[5] = 48,
x[1] + x[2] + 2x[3] + x[4] + x[5] = 24, x[1] + 2x[2] + x[3] + x[4] + x[5] = 12,
2x[1] + x[2] + x[3] + x[4] + x[5] = 6, [x[1], x[2], x[3], x[4], x[5]]);
[[x[1] = −25, x[2] = −19, x[3] = −7, x[4] = 17, x[5] = 65]]
Problem (AIME, 1986/11) The polynomial
1 − x + x2 − x3 + · · · + x16 − x17
may be written in the form
a0 + a1 y + a2 y 2 + · · · + a16 y 16 + a17 y 17 ,
where y = x + 1 and the ai ’s are constants. Find the value of a2 .
Solution: We know that x = y − 1. So
1 − x + x2 + · · · − x17 =
= 1 − (y − 1) + (y − 1)2 − (y − 1)3 + · · · − (y − 1)17 .
We have to find the coefficient of the y 2 term of each power of each binomial, which
78
by the binomial theorem is
17
3
2
.
+ ··· +
+
2
2
2
The Hockey Stick Identity gives that this quantity is equal to
18
= 816.
3
Solution with CAS: f := sum((−1)i xi , i = 0..17);
f := 1−x+x2 −x3 +x4 −x5 +x6 −x7 +x8 −x9 +x10 −x11 +x12 −x13 +x14 −x15 +x16 −x17
taylor(f, x = −1, 18);
18−153(x+1)+816(x+1)2 −3060(x+1)3 +8568(x+1)4 −18564(x+1)5 +31824(x+1)6
−43758(x + 1)7 + 48620(x + 1)8 − 43758(x + 1)9 +
31824(x + 1)10 − 18564(x + 1)11 + 8568(x + 1)12 − 3060(x + 1)13
+816(x + 1)14 − 153(x + 1)15 + 18(x + 1)16 − (x + 1)17
Problem (AIME, 1988/3) Find (log2 x)2 if
log2 (log8 x) = log8 (log2 x).
Solution: Raise both as exponents with base 8,
8log2 (log8 x) = 8log8 (log2 x) .
On the next step we use the definition of logarithms,
23 log2 (log8 x) = log2 x.
79
Applying the fact that k loga x = loga xk , we have
(log8 x)3 = log2 x
On the next step, we use the change of base formula, which states
loga b =
for arbitrary k,
log2 x
log2 8
3
logk b
logk a
= log2 x.
Finally, we have
(log2 x)2 = (log2 8)3 = 27.
Solution with CAS: a := solve(log[2](log[8](x)) = log[8](log[2](x)), x); log[2](a)2 ;
27
Problem (AIME, 1991/1) Find x2 + y 2 if x and y are positive integers such that
xy + x + y = 71
x2 y + xy 2 = 880.
Solution: Let a = x + y and b = xy. Rewriting our original system of eqations we
obtain
a + b = 71
ab = 880.
Solving this system of equations we have a quadratic equation
a2 − 71a + 880 = 0,
80
which factors to
(a − 16)(a − 55) = 0.
Now we get either a = 16 and b = 55 or a = 55 and b = 16. For the first case, it is
easy to see that (x, y) can be (5, 11) (or vice versa). In the second case, since all
factors of 16 must be ≤ 16, no two factors of 16 can sum greater than 32, and so
there are no integral solutions for (x, y). The solution is
52 + 112 = 146.
Solution with CAS: solve(x2 y + xy 2 = 880, xy + x + y = 71, [x, y]);
[[x = 11, y = 5], [x = 5, y = 11],
[x = −RootOf (2Z −55 ∗Z +16) + 55, y = RootOf (2Z −55 ∗Z +16)]]
Problem (AIME, 1994/4) Find the positive integer n for which
⌊log2 1⌋ + ⌊log2 2⌋ + ⌊log2 3⌋ + · · · + ⌊log2 n⌋ = 1994.
(For real x, ⌊x⌋ is the greatest integer ≤ x.)
Solution: It is easy to see that if
2x ≤ a < 2x+1
for some x ∈ Z, then
⌊log2 a⌋ = log2 2x = x.
Thus, there are 2x+1 − 2x = 2x integers a such that
⌊log2 a⌋ = x.
81
Thus the sum of
⌊log2 a⌋
for all such a is x · 2x . Let k be the integer such that
2k ≤ n < 2k+1 .
So for each integer j < k, there are exactly 2j integers a ≤ n such that
⌊log2 a⌋ = j,
and there are n − 2k + 1 such integers with
⌊log2 a⌋ = k.
Therefore,
⌊log2 1⌋ + ⌊log2 2⌋ + ⌊log2 3⌋ + · · · + ⌊log2 n⌋ =
k−1
X
j=0
(j · 2j ) + k(n − 2k + 1) = 1994,
A straightforward computation gives that
7
X
j · 2j = 1538 < 1994
8
X
j · 2j = 3586 > 1994.
j=0
and
j=0
Thus, k = 8 and
k−1
X
(j · 2j ) + k(n − 2k + 1) = 1538 + 8(n − 28 + 1) = 1994
j=0
which implies n = 312.
82
Solution with CAS:
Problem (AIME, 1997/11) Let
P44
cos n◦
x = Pn=1
.
44
◦
n=1 sin n
What is the greatest integer that does not exceed 100x?
Solution: Now we have
P44
cos n◦
cos 1◦ + cos 2◦ + . . . + cos 43◦ + cos 44◦
=
=
x = Pn=1
44
◦
sin 1◦ + sin 2◦ + . . . + sin 43◦ + sin 44◦
n=1 sin n
cos(45 − 1)◦ + cos(45 − 2)◦ + . . . + cos(45 − 43)◦ + cos(45 − 44)◦
=
sin 1◦ + sin 2◦ + . . . + sin 43◦ + sin 44◦
Using the identity
a−b
a+b
cos
sin a + sin b = 2 sin
2
2
we obtain
sin x + cos x = sin x + sin(90 − x) = 2 sin 45 cos(45 − x) =
√
2 cos(45 − x),
so our summation reduces to
1 (cos 1◦ + cos 2◦ + . . . + cos 44◦ ) + (sin 1◦ + sin 2◦ + . . . + sin 44◦ )
x= √ ·
=
sin 1◦ + sin 2◦ + . . . + sin 43◦ + sin 44◦
2
cos 1◦ + cos 2◦ + . . . + cos 43◦ + cos 44◦
1
√ 1+
.
sin 1◦ + sin 2◦ + . . . + sin 43◦ + sin 44◦
2
We can rewrite this equation as
1
x = √ (1 + x).
2
We get x = 1 +
√
2 and ⌊100x⌋ = 241.
83
i),k=1..44
Solution with CAS: evalf 100 cos((1/180)·k·P
;
cos((1/180)·k·P i),k=1..44
241.42135623730950486
Problem (AIME, 1998/5) Given that
Ak =
k(k − 1)π
k(k − 1)
cos
,
2
2
find
|A19 + A20 + · · · + A98 |.
Solution:
k(k − 1)
2
always evaluates to an integer, and the cosine of nπ where n ∈ Z is 1 if n is even
and −1 if n is odd, further, k(k−1)
will be even if 4|k or 4|k − 1, and odd otherwise.
2
So our sum is in the following form:
98
X 19 · 18 20 · 19 21 · 20 22 · 21 23 · 22
Ai = −
+
+
−
−
...,
2
2
2
2
2
i=19
If we group the terms in pairs, we see that we can apply a formula for
n(n − 1) (n + 1)n
n
+
= (n + 1 − (n − 1)) = n.
2
2
2
So the first two fractions add up to 19, the next two to -21, and so forth. If we
pair the terms again now, each pair adds up to −2. There are 98−19+1
= 20 such
2·2
pairs, so the asked sum is | − 2 · 20| = 040.
Solution with CAS: sum(k(k − 1)cos((1/2)k(k − 1)P i), k = 19..98);
−80
84
Problem (AIME, 2000 II/13) The equation
2000x6 + 100x5 + 10x3 + x − 2 = 0
has exactly two real roots, one of which is
√
m+ n
,
r
where m, n and r are integers, m and r are relatively prime, and r > 0. Find
m + n + r.
Solution: We can factor the polynomial on the left hand side:
2000x6 + 100x5 + 10x3 + x − 2 = 2(1000x6 − 1) + x(100x4 + 10x2 + 1) =
= 2[(10x2 )3 − 1] + x[(10x2 )2 + (10x2 ) + 1] =
= 2[10x2 − 1][(10x2 )2 + (10x2 ) + 1] + x[(10x2 )2 + (10x2 ) + 1] =
[(10x2 )2 + (10x2 ) + 1][2(10x2 − 1) + x] =
(20x2 + x − 2)[(10x2 )2 + (10x2 ) + 1].
One can see that
100x4 + 10x2 + 1 ≥ 1 > 0
for every real value of x. Thus the real roots must be the roots of the equation
20x2 + x − 2 = 0.
By the quadratic formula the roots of this are:
x=
So we have
−1 ±
p
√
√
12 − 4(−2)(20)
−1 ± 1 + 160
−1 ± 161
=
=
.
40
40
40
√
−1 + 161
r=
,
40
85
and the final answer is −1 + 161 + 40 = 200.
Remark. A well-known technique for dealing with symmetric (or as in this case,
nearly symmetric) polynomials is to divide through by a power of x with half of
the polynomial’s degree (in this case, divide through by x3 ), and then to use one
of the substitutions t = x ± x1 . In this case, the right substitution
√
1
t = x 10 − √
x 10
gives
t2 + 2 = 10x2 +
and
1
10x2
√
2 10(t3 + 3t) = 200x3 −
2
,
10x3
which reduces the equation to just
√
(t + 3) 2 10t + 1 = 0.
2
Solution with CAS: solve(2000x6 + 100x5 + 10x3 + x − 2 = 0, x);
q
√
1
1
1√
1√
1
− +
161, − −
161, −
−5 + 5I 3,
40 40
40 40
10
q
q
√ 1
√
1
−
−5 − 5I 3,
−5 − 5I 3
10
10
Problem (AIME, 2001, I/3) Find the sum of the roots, real and non-real, of the
equation
2001
1
2001
−x
= 0,
x
+
2
given that there are no multiple roots.
Solution: From Vieta’s formulas, in a polynomial of the form
an xn + an−1 xn−1 + · · · + a0 = 0,
86
the sum of the roots is
−an−1
.
an
From the Binomial Theorem, the first term of
1
−x
2
2001
is −x2001 , so the term with the largest degree is x2000 . Thus we need the coefficient
of that term, as well as the coefficient of x1999 . Using the Binomial Theorem again,
we have
1
1
2001
2001x2000
2000
(−x)
=
1
2
2
and
2
2001 · 2000 1999
1
2001
1999
=−
(−x)
x .
2
8
1
By Vieta’s formulas, we obtain that the sum of the roots is
−
−2001 · 250
2001
2
= 250 · 2 = 500.
Solution with CAS: coeff(x2001 + (1/2 − x)2001 , x2000 );
2001
2
coeff(x2001 + (1/2 − x)2001 , x1999 );
−500250
Problem (AIME, 2006 II/12) Find the sum of the values of x such that
cos3 3x + cos3 5x = 8 cos3 4x cos3 x,
where x is measured in degrees and 100 < x < 200.
87
Solution: We will use the identity
2 cos 4x cos x = cos 5x + cos 3x
by the sum-to-product formulas. Defining a = cos 3x and b = cos 5x, we have
a3 + b3 = (a + b)3 .
From this equation we obtain that ab(a + b) = 0. However,
a + b = 2 cos 4x cos x,
so we get cos x = 0, or cos 3x = 0, or cos 4x = 0, or cos 5x = 0. Hence the solution
set is
A = {150, 126, 162, 198, 112.5, 157.5}
P
and thus x∈A x = 906.
Solution with CAS: See Figure 2.6!
Problem (AIME, 2008 I/8) Find the positive integer n such that
arctan
1
1
1
1
π
+ arctan + arctan + arctan = .
3
4
5
n
4
Solution: Since we are working with acute angles, we have
tan(arctan a) = a.
Note that
tan(arctan a + arctan b) =
by tangent addition. Thus we obtain,
a+b
,
1 − ab
arctan a + arctan b = arctan
a+b
.
1 − ab
88
Figure 2.6: The graph of cos 3x cos 5x(cos 3x + cos 5x)
89
Applying this to the first two terms, we get
arctan
1
1
7
+ arctan = arctan .
3
4
11
arctan
1
23
7
+ arctan = arctan .
11
5
24
Similarly,
We now have
arctan
23
1
π
+ arctan = = arctan 1.
24
n
4
Thus,
+ n1
23 = 1
1 − 24n
23
24
and simplifying, 23n + 24 = 24n − 23, that is n = 047.
Solution with CAS: a:= tan((1/4)pi-arctan(1/3)-arctan(1/4)-arctan(1/5));
(−23 + 24 ∗ tan((1/4)P i))/(24 + 23tan((1/4) ∗ P i))
Problem (AIME, 2011 I/9) Suppose x is in the interval [0, π/2] and
3
log24 sin x (24 cos x) = .
2
Find 24 cot2 x.
Solution: Using the definition of logarithm we can rewrite the given expression as
p
243 sin3 x = 24 cos x.
Square both sides and divide by 242 we get
24 sin3 x = cos2 x.
Using the identity
cos2 x = 1 − sin2 x,
90
we have
24 sin3 x + sin2 x − 1 = 0.
The rational root theorem gives sin x = 13 as a root, and sin−1 13 does fall in the first
quadrant so it satisfies the interval. Thus sin2 x = 19 and, using the Pythagorean
Identity gives us cos2 x = 98 . Then we use the definition of cot2 x to compute our
final answer.
8
cos2 x
2
24 cot x = 24 2 = 24 91 = 24 · 8 = 192.
sin x
9
Solution with CAS: factor(24y 3 + y 2 − 1);
(3y − 1)(8y 2 + 3y + 1)
Problem (AIME, 2013 I/5) The real root of the equation
8x3 − 3x2 − 3x − 1 = 0
can be written in the form
√
3
a+
√
3
b+1
,
c
where a, b, and c are positive integers. Find a + b + c.
Solution: Observe that the equation yields
9x3 = (x + 1)3 ,
so it follows that
√
3
9x = x + 1.
Solving for x yields
so the answer is 98.
1
√
=
3
9−1
√
√
3
81 + 3 9 + 1
,
8
91
Solution with CAS: solve(8x3 − 3x2 − 3x − 1 = 0, x);
3 1/3 1 2/3 1
3 + 3 + ,
8
8
8
...
...
Problem Problem (KöMaL C.826.) Solve the equation
x
3x3
−
= 2.
3
x −1 x−1
Solution: We suppose that x 6= 1 and we have
x
2x3 − x2 − x
3x3
−
=
x3 − 1 x − 1
x3 − 1
which gives
x2 + x − 2 = (x + 2)(x − 1) = 0
The solution of the problem is x = −2.
3
3x
x
Solution with CAS: solve x3 −1 − x−1 = 2, x ;
−2
See Figure 2.7!
Problem (KöMaL K.100.) Solve the following equation in the set of real numbers:
x2 + 6 +
4
1
=
4x
+
.
x2
x
Solution: Let y = x + x1 . Then our equation leads to
y 2 + 4 = 4y
92
Figure 2.7: The graph of
3x3
x3 −1
−
x
x−1
−2
93
Figure 2.8: The graph of x2 + 6 +
1
x2
− 4x −
and we obtain y = 2 és x = 1.
Solution with CAS: See Figure 2.8!
Problem Problem (KöMaL C.881.) Solve the equation
x 2
2
=
x+1
3
3
.
Solution: A straightforward calculation gives
4x3 − 15x2 + 12x + 4 = 0.
4
x
94
Figure 2.9: The graph of
x 2
2
−
x+1 3
3
One can see that x = 2 is a solution of this equation, so
4x3 − 15x2 + 12x + 4 = (x − 2)(4x2 − 7x − 2).
The solutions of the original problem are x1 = 2, x2 = 2, x3 = − 14 .
Solution with CAS: See Figure 2.9!
Problem (KöMaL B.4176.) Solve the equation
(sin x + sin 2x + sin 3x)2 + (cos x + cos 2x + cos 3x)2 = 1.
95
Solution: We will use the identities
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc,
sin α sin β + cos α cos β = cos(β − α)
and we have
3 + 4 cos x + 2 cos 2x = 1.
Since cos 2x = 2 cos2 x − 1, our equation leads to
cos2 x + cos x = 0
This yields cos x = 0 or cos x = −1, that is x =
and l are integer numbers.
π
2
+ kπ or x = π + 2lπ, where k
Solution with CAS: See Figure 2.10!
Problem (KöMaL B.4156.) Solve the equation
π
tg x + ctg x + 1 = cos x +
.
4
Solution: On comparing the range of functions on the left and right hand side we
have
−1 ≥ tg x + ctg x + 1
or
tg x + ctg x + 1 ≥ 3.
Since
π
−1 ≤ cos x +
≤ 1,
4
and the equality is only in the case
π
cos x +
= −1
4
96
Figure 2.10: The graph of cos2 x + cos x
97
Figure 2.11: The graph of 1 + sin x cos x − sin x cos x cos x +
Π
4
which yields x = 3π
+ 2kπ, where k is an integer number. Now the value of the
4
left hand side is −1, so we obtain the solutions of the equation.
Solution with CAS: See Figure 2.11!
Problem (KöMaL B.4157.) Solve the following equation
[x] = x4 − 2x2 .
Solution: We consider the solutions x in the form
x = [x] + {x}
98
where {x} denotes the fraction parts of x. Then we have
{x} = x − x4 − 2x2 ,
and thus
0 ≤ x − x4 − 2x2 < 1.
However, if |x| ≥ 2, then we get
x < 2x2 , x + 2x2 < 4x2 ≤ x4 ,
and −2 < x < 2, and [x] = −2, −1, 0, 1. The corresponding quartic equations are
x4 − 2x2 + 2 = 0,
x4 − 2x2 + 1 = 0,
x4 − 2x2 = 0,
x4 − 2x2 − 1 = 0.
Since
x4 − 2x2 + 2 = (x2 − 1)2 + 1,
there is no real solution of the first equation We can rewrite the second equation
as
(x2 − 1)2 = 0
which yields x = ±1 and, by the condition [x] = −1, we have x = −1. We can
factorize the third equation
x4 − 2x2 = x2 (x2 − 2) = 0.
√
The possible solutions are x = 0, x = ± 2, but the last two values are not solutions
99
Figure 2.12: The graph of [x] and x4 − 2x2
(by the condition [x] = 0). Finally, the fourth equation shows
(x2 − 1)2 = 2
and from this
√
x2 − 1 = ± 2,
√
x2 = 1 + 2,
and
q
√
x = ± 1 + 2.
p
√
Since [x] = 1, we obtain x = 1 + 2.
Solution with CAS: See Figure 2.12!
100
Problem (KöMaL C.973.) Solve the equation
1 + cos 3x = 2 cos 2x.
Solution: It is known that
cos 3x = 4 cos3 x − 3 cos x,
and
cos 2x = 2 cos2 x − 1.
Using these identities we get
4 cos3 x − 4 cos2 x − 3 cos x + 3 = 0.
One can factorize the cubic expression as
4 cos3 x − 4 cos2 x − 3 cos x + 3 = (cos x − 1)(4 cos2 x − 3).
So we have cos x = 1 or cos x = ±
k, l ∈ Z.
√
3
,
2
which gives x1 = 2kπ, x2 = ± π6 + lπ, where
Solution with CAS: solve(1 + cos(3x) − 2cos(2x) = 0);
π 5π
0, ,
6 6
See Figure 2.13!
Problem (KöMaL B.4138.) Solve the equation
√
2(sin x + cos x) = tg x + ctg x.
Solution: For the right side we have
|tg x + ctg x| ≥ 2.
101
Figure 2.13: The graph of 1 + cos 3x − 2 cos 2x
102
However,
√
| 2(sin x + cos x)| = 2 ·
√
2
(| sin x + cos x|) = 2| cos(45◦ − x)| ≤ 2,
2
so all the solutions x of the equation we get
| cos(45◦ − x)| = 1
and x = 45◦ + k · 180◦ . After checking these possible solutions we obtain
x = 45◦ + k · 360◦
as real solutions, where k is an integer number.
Solution with CAS: See Figure 2.14!
Problem (KöMaL C.955.) Solve the equation 10x − 5 = 9[x] in the set of real
numbers (here [x] denotes the integral part of x).
Solution: Let
x = [x] + {x},
where {x} denotes the fractional part of x. Using this notation we have
10[x] + 10{x} − 5 = 9[x]
and
5 − [x] = 10{x}.
It implies that 10{x} is an integer number, so possible values for {x} are k ·0.1, k =
0, 1, 2, . . . , 8, 9. From the previous equation we get [x] and x.
Solution with CAS: See Figure 2.15!
103
Figure 2.14: The graph of
√
2(sin x + cos x) sin x cos x − 1
104
Figure 2.15: The graph of 10x − 9[x] − 5
105
Problem (KöMaL B.4323.) Solve the equation
1 + x4
3
=
4
(1 + x)
4
Solution: Reordering the equation yields
x4 − 12x3 − 18x2 − 12x + 1 = 0
One can observe that this is a symmetric equation and x = 0 is not a solution.
We can divide by x2 and we have
x2 − 12x − 18 −
Introducing the new variable y = x +
1
x
12
1
+ 2 = 0.
x
x
our equation leads to
y 2 − 12y − 20 = 0.
√
Its solutions are y1,2 = 6 ± 56. However, the absolute value of the sum of a
non-zero number and its reciprocal value is at least 2, it is enough to solve the
equation
√
x2 − (6 + 56)x + 1 = 0
The solutions are
x1,2 = 3 +
√
14 ±
q
√
22 + 6 14.
Solution with CAS: See Figures 2.16 and 2.17!
Problem (KöMaL B.4239.) Solve the equation
8x(2x2 − 1)(8x4 − 8x2 + 1) = 1.
Solution: If |x| > 1 then it it easy to see that the absolute value of the product on
the left hand side is at least 8. Thus for all the solutions x we have |x| ≤ 1. There
106
Figure 2.16: The graph of
1+x4
(1+x)4
−
3
4
with x ∈ [0, 3]
107
Figure 2.17: The graph of
1+x4
(1+x)4
−
3
4
with x ∈ [3, 15]
108
exists a real number t for which
x = cos t,
and by using this substitution we get
8 cos t cos 2t cos 4t = 1.
Applying the identity
2 cos α cos β = cos(α + β) + cos(α − β)
we have
2(cos t + cos 3t + cos 5t + cos 7t) − 1 = 0.
It is clear that x = 12 is a solution of the original equation. Applying the trigonometric form of the original equation we obtain the solutions
t2 =
4π
6π
2π
, t3 =
, t4 =
,
7
7
7
π
5π
7π
, t6 =
, t7 =
9
9
9
and the solutions of the original equation are
t5 =
x1 , x2 = cos
2π
4π
6π
, x3 = cos , x4 = cos ,
7
7
7
π
5π
7π
x5 = cos , x6 = cos , x7 = cos .
9
9
9
Since the degree of the original equation is 7 and we have 7 solutions, our problem
is solved.
Solution with CAS: See Figures 2.18 and 2.19!
109
Figure 2.18: The graph of 8x(2x2 − 1)(8x4 − 8x2 + 1) − 1 with x ∈ [−0.4, 0]
110
Figure 2.19: The graph of 8x(2x2 − 1)(8x4 − 8x2 + 1) − 1 with x ∈ [−1, 1]
111
Problem (KöMaL B.4367.) Solve the following equation
3x + 3
x+1
√
.
=4+ √
x
x2 − x + 1
Solution: First we remark that the domain of this equation is the set of positive
real numbers. It is clear that
√
x + 1 ≥ 2 x.
This inequality gives
3x + 3
√
≥ 6,
x
and the equality condition is true if and only if x = 1. We similarly obtain
3(x2 + 1) ≥ 6x,
that is
4(x2 − x + 1) ≥ (x + 1)2 ,
which shows that
4+ √
x+1
≤ 6.
−x+1
x2
The equality holds if and only if x = 1, so the solution of the problem is x = 1.
Solution with CAS: See Figure 2.20!
Problem (KöMaL B.3320.) Solve the equation
xx
1/2
= 1/2.
Solution: We will introduce a new variable, let y = x2 . Rewriting our equation we
have
1
y 2y =
2
One can prove that the function f (y) = y 2y is strictly decreasing on the interval
0 < y ≤ 1e and strictly increasing if y ≥ 1e . This fact gives that we have at most
112
Figure 2.20: The graph of
3x+3
√
x
−4−
√ x+1
x2 −x+1
113
Figure 2.21: The graph of xx
1/2
two solutions and after substitution we obtain exactly two solutions for y,
1
1
and 41 és 16
for x.
4
1
2
and
Solution with CAS: See Figure 2.21!
Problem (KöMaL C.688.) Solve the equation
[x/2] + [x/4] = x.
([x] denotes the integral part of x)
Solution: There is an integer number on the left hand side, so x is an integer
number. The remainder of x dividing by 4 is 0, 1, 2 or 3. In the first case x = 4k,
where k is an integer. Then 2k + k = 4k, which yields k = 0. If x = 4k + 1
then 2k + k = 4k + 1, and k = −1. In the third case x = 4k + 2, now we have
114
Figure 2.22: The graph of [x/2] + [x/4] − x
3k = 4k + 2, k = −2, and finally, x = 4k + 3 gives k = −3. The solutions of the
original problem are x = 0, −3, −6 and −9.
Solution with CAS: See Figure 2.22!
Problem (KöMaL B.3572.) Solve the equation
[x/2] + [x/4] = [x],
where [x] denotes the integral part of x.
Solution: Let [x] = n. Now we have
hxi
2
=
hni
2
,
115
and
hxi
=
hni
4
4
We distinguish four cases. If n = 4k then
.
4k = 2k + k,
and k = 0, n = 0. In the second case n = 4k + 1, then we get
4k + 1 = 2k + k,
and k = −1, n = −3. In the next case n = 4k + 2, we have
4k + 2 = (2k + 1) + k,
and k = −1, n = −2. Finally, if n = 4k + 3, then
4k + 3 = (2k + 1) + k,
and so k = −2, n = −5.
The solutions are real numbers with integral part 0, −2, −3 or −5.
Solution with CAS: See Figure 2.23!
Problem (KöMaL B.3577.) Solve the equation
sin 3x + 3 cos x = 2 sin 2x(sin x + cos x).
Solution: We will introduce new variables. Let
a = cos x, b = sin x.
Using this notation we obtain
sin 2x = 2 sin x cos x = 2ab,
116
Figure 2.23: The graph [x/2] + [x/4] − [x]
117
cos 2x = a2 − b2 = 1 − 2b2 = 2a2 − 1,
sin 3x = sin 2x cos x + cos 2x sin x = 3a2 b − b3 .
Applying the previous formulas our equation leads to
3a2 b − b3 + 3a = 4a2 b + 4ab2 .
This yields
3a = a2 b + 4ab2 + b3 = b(a2 + b2 ) + 4ab2 .
Since a2 + b2 = 1, we have the equations
3a = b + 4ab2 = b + 4a(1 − a2 ),
and
b = a(4a2 − 1).
Taking the square we get
b2 = 1 − a2 = a2 (16a4 − 8a2 + 1).
Now this is a cubic equation (in a2 ), and introducing the new variable A = a2 we
obtain
16A3 − 8A2 + 2A − 1 = (2A − 1)(8A2 + 1) = 0.
This yields A = 12 , a = cos x = ± √12 and x =
π
4
+ kπ, where k is an integer number.
Solution with CAS: Solution with CAS: See Figure 2.24!
Problem (KöMaL B.3579.) Solve the following equation
x=
r
q
√
−3 + 4 −3 + 4 −3 + 4x.
118
Figure 2.24: The graph of sin 3x + 3 cos x − 2 sin 2x(sin x + cos x)
119
Solution: Taking square (several times) we have
x8 + 12x6 + 150x4 + 684x2 − 16384x + 15537 = 0.
We can factorize the polynomial as
x8 + 12x6 + 150x4 + 684x2 − 16384x + 15537 =
(x − 1)(x − 3)(x6 + 4x5 + 25x4 + 88x3 + 427x2 + 1444x + 5179),
thus the solutions of the problem are x1 = 1 and x2 = 3. Further, x ≥ 43 , so the
factor
x6 + 4x5 + 25x4 + 88x3 + 427x2 + 1444x + 5179
is positive for every x.
Solution with CAS: See Figure 2.25!
Problem (KöMaL A.291.) Solve the equation
x=
r
2+
q
2−
√
2 + x.
Solution: It is easy to see that we have to find the solution(s) x in the interval
[−2, 2]. Then there exists an angle y for which 0 ≤ y ≤ π and x = 2 cos y. We
obtain r
r
p
√
y
y
2 + x = 2 + 2 cos y = 2 + 4cos2 − 2 = 2 cos ,
2
2
r
q
π y √
y
y
2 − 2 + x = 2 − 2 cos = 2 sin = 2 cos
−
,
2
4
2 4
and
r
q
π y √
2 + 2 − 2 + x = 2 cos
−
,
4 8
120
Figure 2.25: The graph of x −
q
p
√
−3 + 4 −3 + 4 −3 + 4x
121
q
p
√
Figure 2.26: The graph x − 2 + 2 − 2 + x
that is we have the trigonometric equation
cos y = cos
Its solution is y =
2π
9
π
4
−
y
.
8
and we get x = 2 cos 2π
.
9
Solution with CAS: See Figure 2.26!
Problem (KöMaL B.3536.) Determine the minimum value of the function
x2
+ x cos x + cos 2x.
8
122
Solution: We will apply the identity
cos 2x = 2 cos2 x − 1.
Then we have
x2
x2
1
+ x cos x + cos 2x =
+ x cos x + 2 cos2 x − 1 = (x + 4 cos x)2 − 1 ≥ −1,
8
8
8
and the equality holds if and only if
f (x) = x + 4 cos x = 0.
The value of f at x = 0 is 4, at x = π is π − 4, so there is a point x0 in the interval
[0, π] such that f (x0 ) = 0. Then the minimum value of our function is −1.
Solution with CAS: See Figures 2.27 and 2.28!
Problem (KöMaL B.3524.) The sum of the real numbers x and y is 1. Determine
the maximum value of
xy 4 + x4 y.
Solution: As a preparatory step we factorize our function
xy 4 + x4 y = xy((x + y)3 − 3xy(x + y)) = xy(1 − 3xy).
This is a quadratic function in xy, its maximum point is at xy = 61 , and the value
1
1
. Now we have that the maximum value of the original function is at most 12
is 12
and the possible maximum points are the solution of the system of equations
1
x + y = 1, xy = .
6
An easy calculation gives the maximum points
x1,2 =
1
1
± √ .
2 2 3
123
Figure 2.27: The graph of
x2
8
+ x cos x + cos 2x with −10 < x < 10
124
Figure 2.28: The graph of
x2
8
+ x cos x + cos 2x with −2 < x < 5
125
Figure 2.29: The graph of x(1 − x)4 + x4 (1 − x)
Solution with CAS: See Figure 2.29!
Problem (KöMaL B.4275.) Solve the following equation
√
x6 − x3 − 2x2 − 1 = 2(x − x3 + 1) x.
Solution: It is clear that the domain of the equation is the set of nonnegative real
√
numbers. Let y = x. Using this new variable our equation leads to
y 12 + 2y 7 − y 6 − 2y 4 − 2y 3 − 2y − 1 = 0.
Observe that
y 12 − y 6 − 2y 3 − 1 = y 12 − (y 3 + 1)2 = (y 6 − y 3 − 1)(y 6 + y 3 + 1),
126
and since
2y 7 − 2y 4 − 2y = 2y(y 6 − y 3 − 1),
we can factorize
y 12 + 2y 7 − y 6 − 2y 4 − 2y 3 − 2y − 1 = (y 6 − y 3 − 1)(y 6 + y 3 + 2y + 1) = 0.
Since the summands in the second factor are positive or non-negative it is enough
to deal with the zeros of the first factor. Let z = y 3 , then we have
z2 − z − 1 = 0
and (z is non-negative!), z =
√
1+ 5
,
2
which yields
y=
√ ! 31
1+ 5
,
2
x=
√ ! 23
1+ 5
.
2
and
Solution with CAS:factor (y 12 + 2y 7 − y 6 − 2y 4 − 2y 3 − 2y − 1);
(y 6 − y 3 − 1)(y 6 + y 3 + 2y + 1)
See Figure 2.30!
Problem (KöMaL B.4263.) Solve the following system of equations
x3 + 4y = y 3 + 16x,
1 + x2
= 5.
1 + y2
127
√
Figure 2.30: The graph of x6 − x3 − 2x2 − 1 − 2(x − x3 + 1) x
128
Solution: From the first equation we have
x(x2 − 16) = y(y 2 − 4),
and the second equation gives
y 2 − 4 = 5x2 .
Now we obtain
x(x2 − 16) = 5x2 y.
If x = 0, then y = ±2, otherwise we get
y=
x2 − 16
,
5x
which yields the equation
2
5x + 4 =
x2 − 16
5x
2
.
Introducing the new variable z = x2 we have
124z 2 + 132z − 256 = 0.
We have two solutions of this quadratic equation, however, the first one is negative
and the other one is z = 1. This gives x = ±1 and y = ±3. After checking we get
the solutions
(0, 2), (0, −2), (1, −3), (−1, 3).
Solution with CAS: See Figures 2.31 and 2.32!
Problem (KöMaL C.689.) Solve the equation
2
xlog2 (16x ) − 4xlog2 (4x)+1 − 16log2 (4x)+2 + 64x3 = 0.
129
Figure 2.31: The graph of 5x2 + 4 −
x2 −16
5x
2
with 0.5 ≤ x ≤ 3
130
Figure 2.32: The graph of 5 ∗ x2 + 4 −
x2 −16
5x
2
with −3 ≤ x ≤ −0.5
131
Solution: Our first observation is that x > 0. Divide by 64x3 and introduce the
new variable y = log2 x. Then we have x = 2y , and using this notation the original
equation leads to
2
2
2
22y −y−6 − 2y −4 − 2y −y−2 + 1 = 0.
Now the factorization of the left hand side is
22y
2 −y−6
− 2y
2 −4
− 2y
2 −y−2
+ 1 = (2y
2 −4
− 1)(2y
2 −y−2
− 1) = 0.
and we get
y1 = 1, y2 = −2, y3 = 2,
and
1
x1 = 2, x2 = 4, x3 = .
4
Solution with CAS: See Figures 2.33 and 2.34!
Problem (KöMaL B.4535.) Solve the equation
(x2 − x − 1)2 − x3 = 5.
Solution: The proof is based on the simple observation 5 = 22 + 13 . Reordering
the original equation we get
(x2 − x − 1)2 − 22 = x3 + 1.
After factorization we obtain
(x2 − x − 3)(x2 − x + 1) = (x + 1)(x2 − x + 1),
and
(x2 − x + 1)(x2 − 2x − 4) = 0.
It is easy to see that the first factor (x2 − x + 1 = (x − 21 )2 + 43 of the product
√
is positive for every real x, and the zeros of the second factor are 1 ± 5. These
132
2
Figure 2.33: The graph of xlog2 (16x ) − 4xlog2 (4x)+1 − 16log2 (4x)+2 + 64x3 with 0.2 ≤
x ≤ 0.3
133
2
Figure 2.34: The graph of xlog2 (16x ) − 4xlog2 (4x)+1 − 16log2 (4x)+2 + 64x3 with 0.3 ≤
x ≤ 4.02
134
Figure 2.35: The graph of (x2 − x − 1)2 − x3 − 5
solutions yield the solutions of the original equation.
Solution with CAS: factor((x2 − x + 1)2 − x3 − 5);
(x2 − x + 1)(x2 − 2x − 4)
See Figure 2.35!
Problem (KöMaL B.4467.) Solve the equation
√
x = x2 − 3x + 1 + |x − 1|.
Solution: The domain of our equation is the set of non-negative real numbers. We
135
distinguish two cases according to the sign of x − 1. If x > 1, then we have :
√
x = x2 − 2x,
which gives
x4 − 4x3 + 4x2 − x = 0.
One can factorize the quartic polynomial
x4 − 4x3 + 4x2 − x = x(x − 1)(x2 − 3x + 1),
however, x > 1, so the possible solutions are the zeros of the third factor x2 −3x+1,
that is
√
3± 5
x1,2 =
.
2
After checking we a have
√
3+ 5
2
as a solution with x > 1.
Suppose that x ≤ 1. Then we have
√
x = x2 − 4x + 2
and we get a quartic equation
x4 − 8x3 + 20x2 − 17x + 4 = 0.
It is easy to see that 1 and 4 are zeros of the polynomial, so
x4 − 8x3 + 20x2 − 17x + 4 = (x − 1)(x − 4)(x2 − 3x + 1) = 0.
After short calculation we obtain one solution x =
a solution).
√
3− 5
2
with x ≤ 1 (x = 1 is not
Solution with CAS: See Figure 2.36! solve(x2 − 3x + 1 + |x − 1| −
√
3 1√ 1
5, (−1 + 5)2 .
+
2 2
4
√
x = 0);
136
Figure 2.36: The graph of x2 − 3x + 1 + |x − 1| −
√
x
Problem (KöMaL Gy.3145.) Prove the inequality
1
1
1
1
1
1
7
1
+ 3 + 5 + 7 + 9 + 11 + 13 ≤ 7 .
x x
x
x
x
x
x
x
Solution: Let y = x1 . Now we have
y 13 + y 11 + y 9 − 6y 7 + y 5 + y 3 + y ≤ 0.
One can see that y = 0 and y = 1 are zeros of the polynomial on the left hand
side. So the factorization is
y 13 + y 11 + y 9 − 6y 7 + y 5 + y 3 + y =
137
= y(y − 1)(y 11 + y 10 + 2y 9 + 2y 8 + 3y 7 + 3y 6 − 3y 5 − 3y 4 − 2y 3 − 2y 2 − y − 1),
In the next step we have the factorization of the polynomial of degree 11
y 11 +y 10 +2y 9 +2y 8 +3y 7 +3y 6 −3y 5 −3y 4 −2y 3 −2y 2 −y−1 = (y−1)(y+1)2 (y 8 +3y 6 +6y 4 +3y 2 +1).
Thus we get
y 13 + y 11 + y 9 − 6y 7 + y 5 + y 3 + y = y(y − 1)2 (y + 1)2 (y 8 + 3y 6 + 6y 4 + 3y 2 + 1) ≤ 0.
Since
y 8 + 3y 6 + 6y 4 + 3y 2 + 1
is positive for every y we have the set of solutions is y = 1, y = −1 or y < 0, i. e.
x = 1, x = −1 or x < 0.
Solution with CAS: factor(y 13 + y 11 + y 9 − 6y 7 + y 5 + y 3 + y);
y(y 8 + 3y 6 + 6y 4 + 3y 2 + 1)(y − 1)2 (y + 1)2
Remark: See Figure 2.37!
Problem (KöMaL B.3948.) What is the maximum value of 3a5 b − 40a3 b3 + 48ab5
if a2 + 4b2 = 4?
Solution: Let c = a/2. The our condition is
c2 + b2 = 1,
and
3a5 b − 40a3 b3 + 48ab5 =
96c5 b − 320c3 b3 + 96cb5 = 32cb(3c4 − 10c2 b2 + 3b4 ) =
= 32cb(3(c2 + b2 )2 − 16c2 b2 ) = 32cb(3 − 16c2 b2 ).
Introducing the variable x = 4bc we have to consider the maximum value of the
138
Figure 2.37: The graph of y 13 + y 11 + y 9 − 6y 7 + y 5 + y 3 + y
139
Figure 2.38: The graph of 8x(3 − x2 )
function
8x(3 − x2 ),
where |x| ≤ 2, noting that
±bc ≤ b2 + c2 .
Let
f (x) = 24x − 8x3 .
Now the polynomial may have the maximum point at x = ±2 or at solutions of the
equation f ′ (x) = 0, that is at points x = ±1. The second derivation of f is −48x,
so the function has (local) maximum point at x = 1 and has (local) minimum
point at x = −1.
Solution with CAS: See Figure 2.38!
140
Problem (KöMaL C.542.) Prove the following inequality
(n − 2)(2 + 22 + . . . + 2n ) < n · 2n
in the set of positive integers.
Solution: Using the formula for the sum of geometric progression we obtain
2 + 22 + . . . + 2n = 2n+1 − 2,
so we have to solve the inequality
(n − 2)(2n+1 − 2) < n · 2n .
Expanding the left hand side it follows
n · 2n+1 − 2n+2 − 2n + 4 < n · 2n ,
and
n · 2n + 4 < 2n+2 + 2n
We will apply the well-known inequality 2n ≤ 2n , n = 1, 2, 3, . . . and we have
n · 2n + 4 < 2n+2 + 2n = 5 · 2n ,
and the values n = 1, 2, 3, 4 as possible solutions.
Solution with CAS: See Figure 2.39!
Problem (KöMaL C.873.) What is the maximum point of the function
√
2 sin x − sin x.
Solution: We introduce a new variable. Let
y=
√
2 sin x
141
Figure 2.39: The graph of 2x+2 + 2x − x · 2x − 4
142
and rewriting the original function we have
1
y − y2.
2
The maximum point of this quadratic function is y = 1, so
1
sin x = .
2
Thus we obtain that the maximum points of our function are
π
5π
+ 2kπ,
+ 2lπ
6
6
and this maximum value is 12 .
Solution with CAS: See Figure 2.40!
143
Figure 2.40: The graph of
√
2 sin x − sin x
144
Chapter 3
Other problems
Problem (AIME, 1988/13) Find a if a and b are integers such that x2 − x − 1 is
a factor of ax17 + bx16 + 1.
Solution: Let
F (x) = ax17 + bx16 + 1
and let P (x) be the polynomial such that
P (x)(x2 − x − 1) = F (x).
First, the constant term of P (x) must be −1. Now, we have
(x2 − x − 1)(c1 x15 + c2 x14 + · · · + c15 x − 1),
where c15 is some coefficient. However, since F (x) has no x term, it must be true
that c15 = 1. We have to find c14 now. Notice that all we care about in finding c14
is that
(x2 − x − 1)(· · · + c14 x2 + x − 1) = something + 0x2 + something.
On the other hand, the coefficient of x2 is (−c14 − 1 − 1. Therefore, c14 = −2.
145
146
Following a similar process,
c13 = 3, c12 = −5, c11 = 8,
so the coefficients of P (x) are just the Fibonacci sequence with alternating signs.
Therefore,
a = c1 = F16 ,
where
F16
denotes the 16th Fibonacci number and a = 987.
Solution with CAS: A := solve(x2 − x − 1 = 0, x);
A :=
1 1√ 1 1√
5, −
5
+
2 2
2 2
solve(aA[1]17 + bA[1]16 + 1 = 0, aA[2]17 + bA[2]16 + 1 = 0, [a, b]);
[[a = 987, b = −1597]]
Problem (AIME, 1989) Compute
√
31 · ·30 · 29 · 28 + 1.
Solution: Let us denote our four consecutive integers n − 1, n, n + 1, n + 2. Notice
that
(n − 1)n(n + 1)(n + 2) + 1 = (n2 + n)2 − 2(n2 + n) + 1 = (n2 + n − 1)2 .
Thus, we have
√
31 · ·30 · 29 · 28 + 1 = 869.
147
Solution with CAS: sqrt(n(n + 1)(n + 2)(n + 3) + 1);
p
(1 + 3n + n2 )2
Problem (AIME, 1990/2) Find the value of
√
√
(52 + 6 43)3/2 − (52 − 6 43)3/2 .
√
√
Solution: Suppose that 52 + 6 43 is in the form of (a + b 43)2 . Expanding it we
have
√
√
52 + 6 43 = a2 + 43b2 + 2ab 43.
This implies that a and b equal one of ±1, ±3. The possible pairs are (3, 1) and
(−3, −1), the latter case can be discarded since the square root must be positive.
This means that
√
√
52 + 6 43 = ( 43 + 3)2 .
√
√
Repeating this for 52 − 6 43, the only feasible possibility is ( 43 − 3)2 . Rewriting
the original problem, we get
√
√
( 43 + 3)3 − ( 43 − 3)3 .
Using the difference of cubes, we obtain that
√
√
√
√
[ 43+3 − 43+3] [(43+6 43+9)+(43−9)+(43−6 43+9)] = (6)(3·43+9) = 828.
√
√
Solution with CAS: simplify((52 + 6 43)3/2 − (52 − 6 43)3/2 );
828
Problem (AIME, 1993/5) Let
P0 (x) = x3 + 313x2 − 77x − 8.
148
For integers n ≥ 1, define
Pn (x) = Pn−1 (x − n).
What is the coefficient of x in P20 (x)?
Solution: First notice that
P20 (x) = P19 (x − 20) = P18 ((x − 20) − 19) = P17 (((x − 20) − 19) − 18) . . . .
Applying the formula for the sum of the first n numbers,
1 + 2 + · · · + 20 =
20(20 + 1)
= 210.
2
Therefore, we get
P20 (x) = P0 (x − 210).
On substituting x − 210 into the function definition, we get
P0 (x − 210) = (x − 210)3 + 313(x − 210)2 − 77(x − 210) − 8.
We only need the coefficients of the linear terms, which we can find by the binomial
theorem. (x − 210)3 will have a linear term of
3
2102 x = 630 · 210x.
1
313(x − 210)2 will have a linear term of
2
−313 ·
210x = −626 · 210x,
1
and finally, −77(x − 210) will have a linear term of −77x. Adding up the coefficients, we obtain
630 · 210 − 626 · 210 − 77 = 763.
149
Solution with CAS: expand(subs(x = x − 210, x3 + 313x2 − 77x − 8));
x3 − 317x2 + 763x + 4558462
Problem (AIME, 2000 I/3) In the expansion of (ax + b)2000 , where a and b are
relatively prime positive integers, the coefficients of x2 and x3 are equal. Find
a + b.
Solution: Applying the binomial theorem, we obtain
2000
2b1998 a
=
2000 1997 2
b a
3
which gives b = 666a. Since a and b are positive relatively prime integers, a =
1, b = 666, and a + b = 667.
Solution with CAS: coeff((a ∗ x + b)2000 , x2 );coeff((a ∗ x + b)2000 , x3 );
factor(1331334000b1997 a3 − 1999000b1998 a2 )
Problem (AIME, 2000 II/1) The number
2
3
+
log4 20006 log5 20006
can be written as
m + n.
m
n
where m and n are relatively prime positive integers. Find
Solution: It is easy to see that
3
log4 16
log5 125
2
+
=
+
6
6
6
log4 2000
log5 2000
log4 2000
log5 20006
and
log4 16
lg 16
=
,
6
log4 2000
lg 20006
125
log5 125
=
.
6
log5 2000
lg 20006
150
Thus we get
lg 16
125
lg 2000
1
,
+
=
=
lg 20006 lg 20006
lg 20006
6
and m + n = 7.
Solution with CAS: simplify(2/log[4](20006 ) + 3/log[5](20006 ));
1
6
Problem (AIME, 2000 II/7) Given that
1
1
1
1
1
1
1
1
N
+
+
+
+
+
+
+
=
.
2!17! 3!16! 4!15! 5!14! 6!13! 7!12! 8!11! 9!10!
1!18!
Find the greatest integer that is less than
N
.
100
Solution: The terms of the sum are strongly related to the binomial coefficients.
So, on multiplying by 19! we have
19!
19!
19!
19!
19!
19!
19!
19!N
19!
+
+
+
+
+
+
+
=
,
2!17! 3!16! 4!15! 5!14! 6!13! 7!12! 8!11! 9!10!
1!18!
and
19
19
19
19
19
19
19
19
= 19N.
+
+
+
+
+
+
+
9
8
7
6
5
4
3
2
Recall the identity
19 X
19
n=0
n
= 219
and the property of the binomial coefficients
19
19
=
,
n
19 − n
151
it follows that
9 X
19
n=0
Thus,
n
=
219
= 218 .
2
19
19
19N = 2 −
−
= 218 − 19 − 1 = (29 )2 − 20 = (512)2 − 20 = 262124.
1
0
18
So, N =
262124
19
= 13796 and
N
100
= 137.
Solution with CAS: (1/19)*(sum(binomial(19, i), i = 10..17));
13796
Problem (AIME, 2004, I/7) Let C be the coefficient of x2 in the expansion of the
product
(1 − x)(1 + 2x)(1 − 3x) · · · (1 + 14x)(1 − 15x).
Find |C|.
Solution: Set
P (x) = (1 − x)(1 + 2x)(1 − 3x) · · · (1 + 14x)(1 − 15x).
It is clear that the coefficient of x in P (x) is
−1 + 2 − 3 + . . . + 14 − 15 = −8,
so
P (x) = 1 − 8x + Cx2 + Q(x),
where Q(x) is some polynomial divisible by x3 . Then
P (−x) = 1 + 8x + Cx2 + Q(−x)
152
and so
P (x) · P (−x) = 1 + (2C − 64)x2 + R(x),
where R(x) is some polynomial divisible by x3 . However, we also know
P (x) · P (−x) = (1 − x)(1 + x)(1 + 2x)(1 − 2x) · · · (1 − 15x)(1 + 15x) =
= (1 − x2 )(1 − 4x2 ) · · · (1 − 225x2 ) = 1 − (1 + 4 + . . . + 225)x2 + R(x).
On equating coefficients, we have
2C − 64 = −(1 + 4 + . . . + 225) = −1240,
so −2C = 1176 and |C| = 588.
Solution with CAS: coeff(mul(1 + (−1)i · i · x, i = 1..15), x2 );
−588
Problem (AIME, 2005, I/6) Let P be the product of the non real roots of
x4 − 4x3 + 6x2 − 4x = 2005.
Find ⌊P ⌋.
Solution: It is easy to see that
x4 − 4x3 + 6x2 − 4x + 1 = (x − 1)4 ,
that is our equation lead to
(x + 1)4 − 2006 = 0.
153
One can factorize the polynomial on the left-hand side, which is
((x − 1)2 +
√
2006)((x − 1)2 −
√
2006) = 0.
If you think of each part of the product as a quadratic, then
((x − 1)2 +
√
2006)
is bound to hold the two non-real roots since the other definitely crosses the x-axis
twice since it is just x2 translated down and right. Therefore the products of the
roots of
√
((x − 1)2 + 2006)
is
P =1+
√
2006
so
⌊P ⌋ = 1 + 44 = 45.
Solution with CAS: a:=solve(x4 − 4x3 + 6x2 − 4x − 2005, x);
a := 20061/4 + 1, I · 20061/4 + 1, −20061/4 + 1, −I · 20061/4 + 1
expand(a[2]a[4])
√
2006 + 1
Problem (AIME, 2007, I/3) The complex number z is equal to 9 + bi, where b is
a positive real number and i2 = −1. Given that the imaginary parts of z 2 and z 3
are the same, what is b equal to?
Solution: Squaring, we find that
(9 + bi)2 = 81 + 18bi − b2 .
154
Cubing and ignoring the real parts of the result, we find that
(81 + 18bi − b2 )(9 + bi) = . . . + (9 · 18 + 81)bi − b3 i.
Setting these two equation, we get that
18bi = 243bi − b3 i,
so
b(b2 − 225) = 0
and the possible values for b are −15, 0, 15. Since b > 0, the solution is 15.
Solution with CAS: expand((9 + I · b)3 − (9 + I · b)2 ); factor(225b − b3 )
−b(b − 15)(b + 15)
Problem (AIME, 2007, I/7) Let
N=
1000
X
k=1
k(⌈log√2 k⌉ − ⌊log√2 k⌋).
Find the remainder when N is divided by 1000. (⌊k⌋ is the greatest integer less
than or equal to k, and ⌈k⌉ is the least integer greater than or equal to k.)
Solution: It is easy to see that ceiling of a number minus the floor of a number is
either equal to zero (if the number is an integer),otherwise, it is equal to 1. Thus,
we need to find when or not
log√2 k
is an integer. The change of base formula shows that
log k
2 log k
√ =
.
log 2
log 2
For the log 2 term to cancel out, k is a power of 2. Thus, N is equal to the sum of
155
all the numbers from 1 to 1000, excluding all powers of 2 from 20 = 1 to 29 = 512.
The formula for the sum of an arithmetic sequence and the sum of a geometric
sequence yields that our answer is
(1000 + 1)1000
− (1 + 2 + 22 + . . . + 29 ) (mod 1000).
2
Simplifying, we get
1000
1000 + 1
− 1023
2
(mod 1000) ≡ 500 − 23
(mod 1000) ≡ 477
(mod 1000),
so the answer is 477.
Solution with CAS:
Problem (AIME, 2007, I/8) The polynomial P (x) is cubic. What is the largest
value of k for which the polynomials
Q1 (x) = x2 + (k − 29)x − k
and
Q2 (x) = 2x2 + (2k − 43)x + k
are both factors of P (x)?
Solution: One can see that Q1 and Q2 must have a common zero to both be factors
of the same cubic polynomial. Let this root be a. We then know that a is a root
of
Q2 (x) − 2Q1 (x) = 2x2 + 2kx − 43x + k − 2x2 − 2kx + 58x + 2k = 15x + 3k = 0,
so x =
−k
.
5
We then know that
−k
5
is a root of Q1 so we get that
k2
+ (k − 29)
25
−k
5
−k =0=
= k 2 − 5(k − 29)(k) − 25k = k 2 − 5k 2 + 145k − 25k
156
which implies
k 2 = 30k,
so k = 30 is the highest. We can trivially check into the original equations to find
that k = 30 produces a zero in common, so the answer is 30.
Solution with CAS: resultant(x2 + (k − 29)x − k, 2x2 + (2k − 43)x + k, x);
1080k − 36k 2
factor(1080k − 36k 2 );
−36k(−30 + k)
Problem (AIME, 2008 II/1) Let
N = 1002 + 992 − 982 − 972 + 962 + · · · + 42 + 32 − 22 − 12 ,
where the additions and subtractions alternate in pairs. Find the remainder when
N is divided by 1000.
Solution: Since we have to find the remainder when N is divided by 1000, we may
ignore the 1002 term. Then, applying the difference of squares factorization to
consecutive terms,
(99−98)(99+98)−(97−96)(97+96)+(95−94)(95+94)+· · ·+(3−2)(3+2)−1 =
197
− 193} + |189 {z
− 185} + · · · + |5 {z
− 1} =
| {z
4
=4·
4
4
197 − 5
+ 1 = 100.
8
Solution with CAS: sum((4k)2 + (4k − 1)2 − (4k − 2)2 − (4k − 3)2 , k = 1..25);
10100
157
Problem (AIME, 2010 I/6) Let P (x) be a quadratic polynomial with real coefficients satisfying
x2 − 2x + 2 ≤ P (x) ≤ 2x2 − 4x + 3
for all real numbers x, and suppose P (11) = 181. Find P (16).
Solution: Let
Q(x) = x2 − 2x + 2,
and
R(x) = 2x2 − 4x + 3.
Completing the square, we have
Q(x) = (x − 1)2 + 1,
and
R(x) = 2(x − 1)2 + 1,
so it follows that
P (x) ≥ Q(x) ≥ 1
for all x. Also,
1 = Q(1) ≤ P (1) ≤ R(1) = 1,
so P (1) = 1, and P obtains its minimum at the point (1, 1). Then P (x) must be
of the form c(x − 1)2 + 1 for some constant c, after substituting P (11) = 181 yields
c = 95 . Finally, a straightforward calculation gives
P (16) =
9
· (16 − 1)2 + 1 = 406.
5
Solution with CAS: See Figure 3.1!
Problem (KöMaL B3426.) What is the remainder of the polynomial x2001 dividing
by (x + 1)2 ?
Solution: We know that the remainder is constant or a linear polynomial a · x + b,
158
Figure 3.1: The graphs of x2 − 2x + 2 and 2x2 − 4x + 3
say. Then we can rewrite
x2001 = f (x)(x + 1)2 + a · x + b.
Substituting x = −1-et, we have b − a = −1. After derivation of both sides we get
2001x2000 = f ′ (x)(x + 1)2 + 2(x + 1)f (x) + a,
And substituting x = −1 again we have 2001 = a, and thus the remainder is
2001x + 2000.
Solution with CAS: taylor (x2001 , x = −1, 2002) = −1+2001(x+1)−2001000(x+
1)2 +1333333000(x+1)3 −. . .−1333333000(x+1)1998 +2001000(x+1)1999 −2001(x+
1)2000 + (x + 1)2001 .
Problem (KöMaL B.3446.) Determine the polynomial q(x) such that
p2 (x) − 2p(x)q(x) + q 2 (x) − 4p(x) + 3q(x) + 3 ≡ 0,
where p(x) = x2 + x + 1.
159
Solution: We rewrite the original equation as
(p(x) − q(x))2 ≡ 4p(x) − 3q(x) − 3.
If the degree of q(x) is greater than or equal to 3, the the degree of the polynomial
on the left hand side is 2 deg q, and the degree of the polynomial on the right hand
side is deg q, which is impossible. Now if the degree of q(x) is smaller than 2, the
the right hand side is a quartic and the left hand side is a quadratic polynomial,
we arrive a contradiction again. Thus we have deg q(x) = 2, and let
q(x) = ax2 + (b + 1)x + c + 1.
If the leading coefficient a in not equal to 1, then we have a quartic polynomial
for the right hand side, and a quadratic for the left hand side. so a = 1. In the
sequel we have to find the polynomial q(x) in the form
q(x) = x2 + (b + 1)x + c + 1.
Then
(p(x) − q(x))2 = (bx + c)2 ,
and
4p(x)−3q(x)−3 = 4x2 +4x+4−3x2 −3(b+1)x−3(c+1)−3 = x2 +(1−3b)x−2−3c,
further, on comparing the corresponding coefficients, we get
(b, c) = (1, −1), (−1, −2),
and
q1 (x) = x2 + 2x, q2 (x) = x2 − 1.
Solution with CAS: We are owing the argument above to prove that q(x) is a
160
quadratic polynomial, so q(x) = ax2 +bx+c. p := x2 +x+1; q := a∗x2 +b∗x+c;
p := x2 + x + 1
q := a ∗ x2 + bx + c
collect(expand(p2 − 2pq + q 2 − 4p + 3q + 3), x);
(a2 −2a+1)x4 +(2ab−2b+2−2a)x3 +(−1+2ac−2b−2c+b2 +a)x2 +(−2c−2+b+2bc)x+c2 +c
solve{c2 + c = 0, a2 − 2a + 1 = 0, −2c − 2 + b + 2bc = 0,}
{2ab − 2b + 2 − 2a = 0, −1 + 2ac − 2b − 2c + b2 + a = 0}, [a, b, c];
[[a = 1, b = 0, c = −1], [a = 1, b = 2, c = 0]]
Chapter 4
Exercises
Exercise 1 (IMOLL, 1969) Expand the expression cos π4 + i sin π4
ferent ways, and using these expansions prove that
10
in two dif-
1 10
10
10
= 24 .
−
+
2 5
1
3
Exercise 2 (IMOLL, 1966) How many real solutions are there to the equation
x = 1964 sin x − 189.
Exercise 3 (IMOLL, 1966) Consider the consecutive a1 , a2 , . . . , a99 , a100 positive
integers. What are the last two digits of the sum
a81 + a82 + . . . + a8100 .
Exercise 4 (IMOLL, 1971) It is known that there is a solution (x, y, z) of the
system of equations
x + y + z = 3,
x3 + y 3 + z 3 = 15,
161
162
x4 + y 4 + z 4 = 35
with x2 + y 2 + z 2 < 10. Calculate the sum x5 + y 5 + z 5 .
Exercise 5 (IMOLL, 1972) Determine the solutions of the equation
1 + x + x2 + x3 + x4 = y 2
in the set of integer numbers.
Exercise 6 (IMOLL, 1972) The decimal number 13101 is given. It is instead
written as a ternary number. What are the two last digits of this ternary number?
Exercise 7 (IMOLL, 1973) Is the number
q
3
√
5+2+
q
3
√
5−2
is rational or irrational?
Exercise 8 (IMOLL, 1974) Prove that 2147 − 1 is divisible by 343.
Exercise 9 (IMOLL, 1974) Let n be a positive integer, n ≥ 2, and consider the
polynomial equation
xn − xn−2 − x + 2 = 0.
For each n, determine all complex numbers x that satisfy the equation and have
modulus |x| = 1.
Exercise 10 (IMOLL, 1974) Determine an equation of third degree with integral
coefficients having roots
5π
−3π
π
.
sin , sin , sin
14
14
14
Exercise 11 (IMOLL, 1976) Prove that the number 191976 + 761976 is divisible by
4
the (Fermat) prime number F4 = 22 + 1 and is divisible by at least four distinct
primes other than F4 .
163
Exercise 12 (IMOLL, 1976) Let a, b, c, d be nonnegative real numbers. Prove that
a4 + b4 + c4 + d4 + 2abcd ≥ a2 b2 + a2 c2 + a2 d2 + b2 c2 + b2 d2 + c2 d2 .
Exercise 13 (IMOLL, 1976) Let
x=
√
a+
√
b,
where a and b are natural numbers, x is not an integer, and x < 1976. Prove that
the fractional part of x exceeds 10−19.76 !
Exercise 14 (IMOLL, 1976) Solve the following system of equations:
3x1 − x2 − x3 − x5 = 0,
−x1 + 3x2 − x4 − x6 = 0,
−x1 + 3x3 − x4 − x7 = 0,
−x2 − x3 + 3x4 − x8 = 0,
−x1 + 3x5 − x6 − x7 ,
−x2 − x3 + 3x4 − x8 = 0,
−x1 + 3x5 − x6 − x7 = 0,
−x3 − x5 + 3x7 − x8 = 0,
−x4 − x6 − x7 + 3x8 = 0.
Exercise 15 (IMOLL, 1978) Simplify the following expression:
1
1
1
+
+
,
loga (abc) logb (abc) logc (abc)
where a, b, c are positive real numbers.
164
Exercise 16 (IMOLL, 1978) Find all numbers α for which the equation
x2 − 2x[x] + x − α = 0
has two nonnegative roots. ([x] denotes the largest integer less than or equal to
x.)
Exercise 17 (IMOLL, 1978) A is a 2m-digit positive integer each of whose digits
is 1. B is an m-digit positive integer each of whose digits is 4. Prove that A+B +1
is a perfect square.
Exercise 18 (IMOLL, 1978) If
Cnp =
n!
p!(n − p)!
where p ≥ 1. Prove the identity
p−1
p−1
p−1
Cnp = Cn−1
+ Cn−2
+ . . . + Cpp−1 + Cp−1
,
and then evaluate the sum
1 · 2 · 3 + 2 · 3 · 4 + . . . + 97 · 98 · 99.
Exercise 19 (BW, 2010) For which k do there exist k pairwise distinct primes
p1 , p2 , . . . pk such that
p21 + p22 + . . . + p2k = 2010?
Exercise 20 (IMOLL, 1979) Prove the following inequality:
20
21
< sin 20◦ <
60
60
165
Exercise 21 (IMOLL, 1979) Prove that
1 p
· 4 sin2 36◦ − 1 = cos 72◦
2
Exercise 22 (BW, 1979) Find all triples (a, b, c) of integers satisfying
a2 + b2 + c2 = 20122012.
Exercise 23 (BW, 2008) How many pairs (m, n) of positive integers with m < n
fulfill the equation
3
1
1
=
+ ?
2008
m n
Exercise 24 (BW, 2006) A 12-digit positive integer consisting only of digits 1, 5
and 9 is divisible by 37. Prove that the sum of its digits is not equal to 76.
Exercise 25 (INMO (Second Round), 1986) Let f be a function such that
1
(x2 − 2x + 1) sin x−1
.
f (x) =
sin πx
Find the limit of f in the point x0 = 1.
Exercise 26 (IMOLL, 1979) If p and q are positive integers with
p
1 1 1
1
1
= 1 − + − + ... −
+
q
2 3 4
1318 1319
then prove that p is divisible by 1979.
Exercise 27 (IMOLL, 1979) Prove that
1p
4 · sin2 36◦ − 1 = cos 72◦ .
2
166
Exercise 28 (IMOLL, 1980) Solve the equation
x3 + x2 y” + xy 2 + y 3 = 8(x2 + xy + y 2 + 1),
where x and y are unknown real numbers.
Exercise 29 (INMO (Second Round), 1986) Prove that
arctan
1
π
1
+ arctan = .
2
3
4
Exercise 30 (IMOLL, 1982) Solve the equation
x3 − y 3 = 2xy + 8
in the set of integer numbers.
Exercise 31 (IMOLL, 1982) Determine all real values of the parameter a for which
the equation
16x4 − ax3 + (2a + 17)x2 − ax + 16 = 0
has exactly four distinct real roots that form a geometric progression.
Exercise 32 (IMOLL, 1982) Let p(x) be a cubic polynomial with integer coefficients with leading coefficient 1 and with one of its roots equal to the product of
the other two. Show that 2p(−1) is a multiple of
p(1) + p(−1) − 2(1 + p(0)).
Exercise 33 (IMOLL, 1983) If α is the real root of the equation E(x) = x3 − 5x −
50 = 0 such that xn+1 = (5xn + 50)1/3 and x1 = 5, where n is a positive integer,
prove that
x3n+1 − α3 = 5(xn − α)
167
and
α < xn+1 < xn .
Exercise 34 (IMOLL, 1983) Let (Fn )n≥1 be the Fibonacci sequence
F1 = F2 = 1, Fn+2 = Fn+1 + Fn (n ≥ 1),
and P (x) the polynomial of degree 990 satisfying
P (k) = Fk ,
for k = 992, . . . , 1982. Prove that
P (1983) = F1983 − 1.
Exercise 35 (IMOLL, 1983) Which of the numbers 1, 2, . . . , 1983 has the largest
number of divisors?
Exercise 36 (IMOLL, 1984) Prove that the product of five consecutive positive
integers cannot be the square of an integer.
Exercise 37 (IMOLL, 1984) Let a, b, c be positive numbers with
√
3
a+b+c=
2
Prove that the system of equations
√
y−a+
√
z − a = 1,
√
z − b + x − b = 1,
√
√
x−c+ y−c=1
√
168
has exactly one solution (x, y, z) in real numbers.
Exercise 38 (IMOLL, 1984) Consider all the sums of the form
1985
X
εk k 5
k=1
where ε ∈ {−1, 1}. What is the smallest nonnegative value attained by a sum of
this type?
Exercise 39 (IMOLL, 1985) Solve the following system of equations
√
x−
x+
1
− 2w + 3z = 1,
y
1
− 4w2 − 9z 2 = 3,
y2
√
1
x x − 3 − 8w3 + 27z 3 = −5,
y
x2 +
1
− 16w4 − 81z 4 = 15.
4
y
Exercise 40 (IMOLL, 1985) Solve the following equation in the set of positive
integers
4
1 1 1
+ + = .
x y z
5
Exercise 41 (IMOLL, 1985) Are there integers m and n such that
5m2 − 6mn + 7n2 = 1985.
Exercise 42 (IMOLL, 1985) Let a, b, and c be real numbers such that
1
1
1
+
+
= 0.
bc − a2 ca − b2 ab − c2
169
Prove that
a
b
c
+
+
= 0.
2
2
2
2
(bc − a )
(ca − b )
(ab − c2 )2
Exercise 43 (IMOLL, 1985) Factorize 51985 − 1 as a product of three integers,
each greater than 5100 .
Exercise 44 (IMOLL, 1985) Let n ≥ 1 be a positive integer and
An =
n
X
k6
k=1
2k
.
Evaluate
lim An .
n→∞
Exercise 45 (IMOLL, 1985) Let x, y, and z be real numbers satisfying
x + y + z = xyz.
Prove that
x(1 − y 2 )(1 − z 2 ) + y(1 − z 2 )(1 − x2 ) + z(1 − x2 )(1 − y 2 ) = 4xyz.
Exercise 46 (IMOLL, 1986) Find the last eight digits of the binary development
of 271986 .
Exercise 47 (IMOLL, 1986) Find four positive integers each not exceeding 70000
and each having more than 100 divisors.
Exercise 48 (IMOLL, 1986) Prove the inequality
(−a + b + c)2 (a − b + c)2 (a + b − c)2 ≥ (−a2 + b2 + c2 )(a2 − b2 + c2 )(a2 + b2 − c2 )
for all a, b, c.
170
Exercise 49 (IMOLL, 1986) Find all integers x, y, z such that
x3 + y 3 + z 3 = x + y + z = 8.
Exercise 50 (IMOLL, 1987) Determine the least possible value of the natural
number n such that n! ends in exactly 1987 zeros.
Exercise 51 (IMOLL, 1987) Find, with argument, the integer solutions of the
equation
3z 2 = 2x3 + 385x2 + 256x − 58195.
Exercise 52 (IMOLL, 1988) Let n be a positive integer. Find the number of odd
coefficients of the polynomial
un (x) = (x2 + x + 1)n .
Exercise 53 (IMOLL, 1988) Let a be the greatest positive root of the equation
x3 − 3 · x2 + 1 = 0. Show that [a1788 ] and [a1988 ] are both divisible by 17. Here [x]
denotes the integer part of x.
Exercise 54 (IMOLL, 1988) Let
g(x) = x5 + x4 + x3 + x2 + x + 1.
What is the remainder when the polynomial g(x12 ) is divided by the polynomial
g(x)?
Exercise 55 (IMOLL, 1988) Find all positive integers x such that the product of
all digits of x is given by x2 − 10 · x − 22.
Exercise 56 (IMOLL, 1988) Show that the solution set of the inequality
70
X
k=1
k
5
≥
x−k
4
171
is a union of disjoint intervals, the sum of whose length is 1988.
Exercise 57 (IMOLL, 1988) (i) Calculate x if
x=
(11 + 6 ·
p
p
√
√
√
2) · 11 − 6 · 2 − (11 − 6 · 2) · 11 + 6 · 2
p√
p√
p√
(
5+2+
5 − 2) − (
5 + 1)
√
(ii) For each positive number x, let
k=
x+
x
Calculate the minimum value of k.
1 6
− x6 + x16 −
x
3
+ x1 − x3 + x13
2
Exercise 58 (IMOLL, 1988) The Fibonacci sequence is defined by
an+1 = an + an−1 , n ≥ 1, a0 = 0, a1 = a2 = 1.
Find the greatest common divisor of the 1960-th and 1988-th terms of the Fibonacci sequence.
Exercise 59 (IMOLL, 1989) It is known that y
cos x + cos y + cos z
sin x + sin y + sin z
=
= a.
cos(x + y + z
sin(x + y + z)
Prove that
cos(y + z) + cos(x + z) + cos(x + y) = a.
Exercise 60 (IMOLL, 1990) Find the minimal value of the function
f (x) =
√
q
q
q
√
√
√
15 − 12 cos x+ 4 − 2 3 sin x+ 7 − 4 3 sin x+ 10 − 4 3 sin x − 6 cos x
Exercise 61 (IMOLL, 1980) Find the digits left and right of the decimal point in
172
the decimal form of the number
√
√
( 2 + 3)1980 .
Exercise 62 (IMOLL, 1980) Prove that
tan 730′ =
√
6+
√
2−
√
3 − 2.
Exercise 63 (IMOLL, 1989) Solve the equation
3x3 − [x] = 3
in the set of real numbers.
Exercise 64 (IMOLL, 1989) A sequence of real numbers x0 , x1 , x2 , . . . is defined
as follows: x0 = 1989 and for each n ≥ 1
n−1
xn = −
Calculate the value of
1989 X
xk .
n k=0
1989
X
2n xn .
n=0
Exercise 65 (IMOLL, 1989) Let
f (x) = a sin2 x + b sin x + c,
where a, b, and c are real numbers. Find all values of a, b and c such that the
following three conditions are satisfied simultaneously:
(i) f (x) = 381 if sin x = 21 .
(ii) The absolute maximum of f (x) is 444.
(iii) The absolute minimum of f (x) is 364.
173
Exercise 66 (IMOLL, 1989) Let S1 , S2 be squares. Solve the equation S2 − S1 =
1989.
Exercise 67 (IMOLL, 1989) Prove the following identity
159
X
1 2 1 1 2
641
1
1
2
1 + − + + − + ... +
+
−
=2
.
2 3 4 5 6
478 479 480
(161
+
k)(480
−
k)
k=0
Exercise 68 (IMOLL, 1990) Prove that the number
√
2+
√
3+
√
1990
is irrational.
Exercise 69 (IMOLL, 1990) Prove the following equality
995
X
(−1)k
1
1991 − k
=
.
k
1991
−
k
1991
k=0
Exercise 70 (IMOLL, 1990) Let f (0) = f (1) = 0 and
2
f (n + 2) = 4n+2 f (n + 1) − 16n+1 f (n) + n · 2n , n = 0, 1, 2, . . . .
Prove that f (1989), f (1990) és f (1991) are divisible by 13.
Exercise 71 (IMOLL, 1990) Let a, b, c be arbitrary real numbers. Prove that
(a2 + ab + b2 )(b2 + bc + c2 )(c2 + ac + a2 ) ≥ (ab + ac + bc)3 .
Exercise 72 (IMOLL, 1990) Solve the following system of equations
x3 + y 3 = 1,
174
x5 + y 5 = 1
in the set of real numbers.
Exercise 73 (IMOLL, 1990) What is the minimum value of the expression
√
15 − 12 cos x +
q
√
4 − 2 3 sin x +
q
q
√
√
7 − 4 3 sin x + 10 − 4 3 sin x − 6 cos x.
Exercise 74 (IMOSL, 2009) Let a, b, c be positive numbers with
1 1 1
+ + = a + b + c.
a b c
Prove that
1
1
1
3
+
+
≤ .
2
2
2
(2a + b + c)
(a + 2b + c)
(a + b + 2c)
16
Exercise 75 (IMOSL, 1996) Let a, b, c be positive numbers with abc = 1. Prove
that
bc
ca
ab
+
+
≤ 1.
5
5
5
5
ab + a + b
bc + b + c
ac + a5 + c5
Exercise 76 (IMOSL, 2000) Let a, b, c be positive real numbers with abc = 1.
Prove that
1
1
1
b−1+
c−1+
≤ 1.
a−1
b
c
a
Exercise 77 (IMOSL, 1975) Let x0 = 5 and xn+1 = xn +
1
.
xn
Prove that
45 < x1000 < 45.1.
Exercise 78 (IMOSL, 1995) Let a, b, c be positive real
Show that
1
1
1
+ 3
+ 3
≥
3
a (b + c) b (a + c) c (a + b)
numbers with abc = 1.
3
.
2
175
Exercise 79 (IMOSL, 1975) Determine
" 109
X
#
n−2/3 .
n=1
Exercise 80 (IMOSL, 1960) Solve the following inequality
4x2
√
< 2x + 9.
(1 − 2x + 1)2
Exercise 81 (IMOSL, 1960) Determine all three-digit numbers N having the
N
property that N is divisible by 11, and 11
is equal to the sum of the squares of the
digits of N .
Exercise 82 (IMOSL, 1967) Solve the following system of equations
x2 + x − 1 = y,
y 2 + y − 1 = z,
z 2 + z − 1 = x.
Exercise 83 (IMOSL, 1967) Solve the following system of equations
|x + y| + |1 − x| = 6,
|x + y + 1| + |1 − y| = 4.
Exercise 84 (IMOSL, 1967) Prove that
cos x < 1 −
for every real x ∈ 0, π2
x2 x4
+
2
16
176
Exercise 85 (JMOP, 2012) Determine the positive integers n for which the product of their divisors is 24240 .
Exercise 86 (JMOP, 2012) How many positive integers n are there such that
1000000
1000000
−
= 1.
n
n+1
Exercise 87 (TITEE, 2012) Find the number of digits of
99
X
3n .
n=0
Exercise 88 (TITEE, 2012) For real number a denote by [a] the greatest integer
√
not exceeding a. How many positive integers n ≤ 10000 are there which is [ n] is
a divisor of n?
Exercise 89 (TITEE, 2012) Let n be positive integer. Define a sequence {ak } by
k
1
1
nX
a1 =
, ak+1 = −
+
ai (k = 1, 2, 3, · · ·).
n(n + 1)
k + n + 1 k i=1
Let
bn =
n
X
√
ak .
k=1
Prove that
lim bn = ln 2.
n→∞
Exercise 90 (FTST, 2006) Let a, b, c be positive real numbers with abc = 1. Prove
that
a
b
c
3
+
+
≥ .
(a + 1)(b + 1) (b + 1)(c + 1) (c + 1)(a + 1)
4
177
Exercise 91 (ARO, 2009) Let a, b, c real numbers with
(a + b)(b + c)(c + a) = abc,
and
(a3 + b3 )(b3 + c3 )(c3 + a3 ) = a3 b3 c3 .
Prove that abc = 0.
Exercise 92 (IZO, 2005) Let a, b, c be positive real numbers. Prove that
a
b
c
+
+
≥ 1.
a + 2b b + 2c c + 2a
Exercise 93 (IZO, 2005) Let a, b, c, d be positive real numbers. Prove that
d
a
b
4
c
+
+
+
≥ .
a + 2b b + 2c c + 2d d + 2a
3
Exercise 94 (IZO, 2006) Let a, b, c, d be real numbers with
a+b+c+d=0
Prove that
(ab + ac + ad + bc + bd + cd)2 + 12 ≥ 6(abc + abd + acd + bcd).
Exercise 95 (IZO, 2006) Solve in positive integers the equation
n = ϕ(n) + 402,
where ϕ(n) is the number of positive integers less than n having no common prime
factors with n.
178
Exercise 96 (RNO, 2001) Show that there exist no integers a and b such that
a3 + a2 b + ab2 + b3 = 2001.
Exercise 97 (RNO, 2001) Solve the equation
2x
2 +x
+ log2 x = 2x+1
in the set of real numbers.
Exercise 98 (USAIMTS, 1992) For an arbitrary positive integer n let P (n) denote
the number of its divisors. Determine the smallest n with
P (P (P (n))) > 1012 .
Exercise 99 (HMMT, 2008) Let x, y be positive real numbers with
x2 + y 2 = 1, x4 + y 4 =
17
.
18
Determine xy.
Exercise 100 (HMMT, 2008) Let f (x) = x3 + x + 1. Suppose g is a cubic
polynomial such that g(0) = −1, and the roots of g are the squares of the roots of
f . Find g(9).
Exercise 101 (HMMT, 2008) Evaluate:
∞ X
n−1
X
k
.
n+k
2
n=1 k=1
Exercise 102 (HMMT, 2008) Find p so that
lim x
x→∞
p
√
3
x+1+
√
3
√
3
x−1−2 x
179
is some non-zero real number.
Exercise 103 (HMMT, 2008) Let
f (x) = sin6
x
4
+ cos6
x
4
.
for all real numbers x. Determine f (2008) (0) (i.e., f differentiated 2008 times and
then evaluated at x = 0).
Exercise 104 (HMMT, 2008) Let
T =
Z
0
ln 2
2e3x + e2x − 1
dx.
e3x + e2x − ex + 1
Determine exp{T }.
Exercise 105 (HMMT, 2008) Determine all pairs (a, b) of real numbers such that
10, a, b, ab
is an arithmetic progression.
Exercise 106 (HMMT, 2008) Find the real solution(s) to the equation
(x + y)2 = (x + 1)(y − 1).
Exercise 107 (HMMT, 2008) Given that
x + sin y = 2008,
and
x + 2008 cos y = 2007,
where 0 ≤ y ≤ π2 . Find the value of x + y.
180
Exercise 108 (HMMT, 2008) Determine
∞
X
n=1
n4
n
.
+4
Exercise 109 (HMMT, 2008) Solve the following equation
v
s
u
r
u
q
√
t
√
x + 4x + 16x + . . . + 42008 x + 3 − x = 1
in the set of real numbers.
Exercise 110 (HMMT, 2008) Let P (x) be a polynomial with degree 2008 and
leading coefficient 1 such that
P (0) = 2007, P (1) = 2006, P (2) = 2005, . . . , P (2007) = 0.
Determine the value of P (2008). You may use factorials in your answer.
Exercise 111 (HMMT, 2008) Let (x, y) be a pair of real numbers satisfying
56x + 33y = −
y
,
x2 + y 2
33x − 56y =
x
.
+ y2
x2
Determine the value of |x| + |y|.
Exercise 112 (HMMT, 2008) Let a, b, c be nonzero real numbers such that
a + b + c = 0, a3 + b3 + c3 = a5 + b5 + c5 .
Find the value of a2 + b2 + c2 ?
181
Exercise 113 (HMMT, 2008) Compute
arctg (tg 65◦ − 2 tg 40◦ )?
Exercise 114 (HMMT, 2008) Evaluate:
∞ X
2n 1
.
n 5n
n=0
Exercise 115 (HMMT, 2004) Find the largest number n such that (2004!)! is
divisible by ((n!)!)!.
Exercise 116 (HMMT, 2004) Compute
20033
20053
−
.
2003 · 2004 2004 · 2005
Exercise 117 (HMMT, 2004) Solve the equation
x4 + (2 − x)4 = 34
in the set of real numbers.
Exercise 118 (HMMT, 2004) Let x be a real number such that
x3 + 4x = 8.
Determine the value of x7 + 64x2 .
Exercise 119 (HMMT, 2004) There exists a polynomial P of degree 5 with the
following property: if z is a complex number such that
z 5 + 2004z = 1,
182
then P (z 2 ) = 0. Calculate the quotient
P (1)
.
P (−1)
Exercise 120 (HMMT, 2004) Let k, x, y be positive real numbers with
3=k
2
x2 y 2
+
y 2 x2
+k
x y
+
y x
.
What is the maximum value of k.
Exercise 121 (SMT, 2010) How many zeros are there at the end of
200
124
?
Exercise 122 (SMT, 2010) What is the sum of the solutions to the following
equation
1
2
3
4
+ 2
+ 2
+ 2
= 2010x − 4.
2
x −1 x −2 x −3 x −4
Exercise 123 (SMT, 2006) Find the smallest nonnegative integer n for which
2006
n
is divisible by 73 .
Exercise 124 (SMT, 2006) Let a, b, c be real numbers satisfying:
ab − a = b + 119,
bc − b = c + 59,
ca − c = a + 71.
Determine all possible values of a + b + c.
183
Exercise 125 (SMT, 2006) Let a and b be non zero digits. Solve the equation
aabb = n4 − 6n3 .
Exercise 126 (SMT, 2006) Evaluate:
∞
X
k=1
1
√
k k + 2 + (k + 2) k
√
Exercise 127 (SMT, 2006) Evaluate:
∞
X
arctg
n=1
1
n2 − n + 1
Exercise 128 (SMT, 2006) What is the minimum value of the expression
2x2 + 2y 2 + 5z 2 − 2xy − 4yz − 4x − 2z + 15
where x, y, z are real numbers.
Exercise 129 (SMT, 2006) Determine the following sum
10
X
x=2
1
.
− 1)
x(x2
Exercise 130 (PUMC, 2009) What is the common zero of the following three
polynomials?
x3 + 41x2 − 49x − 2009,
x3 + 5x2 − 49x − 245,
x3 + 39x2 − 117x − 1435.
184
Exercise 131 (PUMC, 2009) For what positive integer k is
100
200
·
k
k
maximal?
Exercise 132 (PUMC, 2009) You are given that
17! = 355687ab809600
for some digits a and b. Find the two-digit number ab that is missing above.
Exercise 133 (PUMC, 2010) Find the sum of the coefficients of the polynomial
(63x − 61)4 .
Exercise 134 (PUMC, 2010) Calculate
∞ h
X
√
n
n=1
i
2010 − 1 ,
where ⌊x⌋ is the largest integer less than or equal to x.
Exercise 135 (PUMC, 2010) Let S be the sum of all real x such that
4x = x4 .
Find the nearest integer to S.
Exercise 136 (PUMC, 2010) Define
f (x) = x +
s
x+
r
q
√
x + x + x + . . ..
185
Find the smallest integer x such that
√
f (x) ≥ 50 x.
Exercise 137 (PUMC, 2010) Let α, β, γ denote the roots of the equation
3x3 − 5x2 + 2x − 6 = 0.
Calculate the expression
1
α−2
2
+
1
β−2
2
+
1
γ−2
2
.
Exercise 138 (PUMC, 2010) Let
√
5
√
√
√
√
1
5
5
5
5
= a + b 2 + c 4 + d 8 + e 16,
2−1
where a, b, c, d, e are integer numbers. Determine a2 + b2 + c2 + d2 + e2 ?
Exercise 139 (PUMC, 2010) Find the largest positive integer n such that
σ(n) = 28,
where σ(n) is the sum of the divisors of n, including n.
Exercise 140 (PUMC, 2010) Simplify the following fraction
2011digitones
z }| {
1010 11 . . . 11 0101
.
1100 |11 .{z
. . 11} 0011
2011ones
186
Exercise 141 (PUMC, 2010) Let f (n) be the sum of the digits of n. Find
99
X
f (n).
n=1
Exercise 142 (AIME, 1984) Determine ab, if ha
log8 a + log4 b2 = 5
and
log8 b + log4 a2 = 7?
Exercise 143 (KöMaL B.3740.) Let
2an
1
− an−1 , (n ≥ 1).
a0 = 1, a1 = , an+1 =
3
3
Prove that there is positive integer n for which an > 0, 9999.
Exercise 144 (KöMaL B.3451.) Solve the following equation
arcsin x + arcsin
√
15x =
π
.
2
Exercise 145 (KöMaL B.3863.) Prove that the number
12005 − 22006 + 32006 − 42006 + . . . − 20042006 + 20052006
is divisible by 1003.
Exercise 146 (KöMaL B.3846.) Solve the following equation
√
4
2−x+
√
4
15 + x = 3.
187
Exercise 147 (KöMaL C.1048.) Show that
2 cos 40◦ − cos 20◦ √
= 3.
sin 20◦
Exercise 148 (KöMaL K.267.) We know that
ab + acb = 2 · ba,
where ab, ba are two-digit numbers and acb is a three-digit number. Determine the
digits a and b if c = 0.
Exercise 149 (KöMaL C.729.) Solve the equation
2x lg x + x − 1 = 0
in the set of real numbers.
Exercise 150 (KöMal B.4531.) Solve the equation
(x2 + 100)2 = (x3 − 100)3 .
Exercise 151 (KöMaL A.358.) Let a, b, c be positive numbers with abc = 1.
Prove the foillowing inequality
1 1 1
3
+ + −
≥3
a b c a+b+c
1
1
1
+ 2+ 2
2
a
b
c
·
a2
1
.
+ b2 + c 2
Exercise 152 (KöMaL Gy.3175.) Let a, b be positive real number. Prove the
following inequality
(a − b)2
≤
2(a + b)
r
(a − b)2
a2 + b 2 √
√
.
− ab ≤
2
2(a + b)
188
Exercise 153 (KöMaL B.4556.) Solve the system of equations
x3 = 5x + y,
y 3 = 5y + x.
Chapter 5
Some useful theorems
Theorem 1 (Binomial theorem) Let a, b real numbers and let n be a positive
integer. Then we have
n n
n
n n−2 2
n n−1
n n
n−1
b
ab
+
a b + ... +
a b+
a +
(a + b) =
n
n−1
2
1
0
n
Proof. See [2].
Theorem 2 (Euler-Fermat Theorem) If n and a are coprime positive integers,
then
aϕ(n) ≡ 1 (mod n)
where ϕ(n) is the Euler’s totient function.
Proof. See [1].
Theorem 3 (Wilson’s Theoewm) Let p be a prime number. Then we have
(p − 1)! ≡ −1
Proof. See [1].
189
(mod p).
190
Theorem 4 (Moivre’s formula) Let n be a positive integer. If the complex number
z = |z|(cos α + i sin α)
then we get
z n = |z|n (cos nα + i sin nα).
Proof. See [2].
Theorem 5 (Legendre’s formula) Let p be a prime number and let ordp (n) denote
the p-order of n. Then we get
ordp (n!) =
∞ X
n
i=1
Proof. See [1].
pi
.
Bibliography
[1] Gyarmati Edit és Turán Pál, Számelmélet, Tankönyvkiadó, Budapest,
1989.
[2] Szendrei János, Algebra és számelmélet, Nemzeti Tankönyvkiadó, Budapest, 1996.
[3] www.komal.hu.
[4] www.artofproblemsolving.com.
191
Contents
1 Algebra and Number Theory . . . . . . . . . . . . . . . . . . . . .
2 Equations and Inequalities
5
. . . . . . . . . . . . . . . . . . . . . . 63
3 Other problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
5 Some useful theorems
. . . . . . . . . . . . . . . . . . . . . . . . . 189
192