Download Math 117 J. Basilla Institute of Mathematics University of the

Institute of Mathematics
University of the Philippines-Diliman
Math 117
J. Basilla
Review
1 December 2010
1. Denote the greatest integer function by bxc.
(a) Show that the number of positive
integers less than or equal to x that are divisible by the
positive integer d is given by xd .
Solution: The numbers divisible by d are d, 2d, 3d, . . . , qd, where qd ≤ x < qd + d. This
gives us q positive numbers. We now express q in terms of n and d. Since qd ≤ x < qd + d,
we have x = qd + r, where 0 ≤ r < d. Therefore, xd = q + dr , where 0 ≤ dr < 1. Hence,
b xd c = q.
(b) How many integers between 100 and 1000 are divisible by 7? by 49?
Solution: The largest number divisible by 7 which do not exceed 100 is 98 = 14 ∗ 7. The
largest number divisible by 7 which do not exceed 1000 is 994 = (142) ∗ 7. Therefore, there
are 128 = 142 − 14 numbers divisible by 7.
100
Equivalently, the number we are looking for is b 1000
7 c − b 7 c = 142-14 = 128.
The largest number divisible by 49 which do not exceed 100 is 98 = 2 ∗ 49. The largest
number divisible by 49 which do not exceed 1000 is 980 = (20) ∗ 49. Therefore, there are
18 = 20 − 2 numbers divisible by 7.
100
Equivalently, the number we are looking for is b 1000
49 c − b 49 c =20-2 = 18.
(c) Find the number of positive integers not exceeding 1000 that are divisible by 5, by 25, by 125,
and by 625.
1000
1000
Solution: The answers are respectively b 1000
5 c = 200, b 25 c = 40, b 125 c = 8, and
1000
b 625 c = 1.
(d) If 2010! is expanded and is written in base 2, in how many zeros will it end?
Solution: The problem asks for the highest power of 2 which divides 2010!. That is, find r
such that 2r k2010!. We need to count the number of 2 in the prime factorization of 2010!.
Each even number contributes a 2 and there are b 2010
2 c even numbers from 1 to 2010, hence,
we expect to have at least b 2010
c
in
the
prime
factorization
of 2010!.
2
Every multiple of 4 contributes at least two 2’s. The first contribution was already counted
when we counted these multiples of 4 as even. Hence, there are b 2010
4 c, additional 2’s in
the prime factorization of 2010!.
Every multiple of 8 contributes at least three 2’s. The first contribution was already counted
when we counted these multiples of 8 as multiples of 4. Hence, there are b 2010
8 c, additional
2’s in the prime factorization of 2010!.
We continue in the same manner. Hence, the number we are looking for is
∞
X
2010
b i c
2
i=1
= b
2010
2010
2010
2010
2010
2010
2010
c+b
c+b
c+b
c+b
c+b
c+b
c
2
4
8
16
32
64
128
=
2010
2010
2010
2010
c+b
c+b
c+b
c + ···
256
512
1024
2048
1000 + 502 + 251 + 125 + 62 + 31 + 15 + 7 + 3 + 1 + 0 + 0 + · · ·
=
2002.
+b
(e) If 2010! is expanded and is written the usual way( base 10), in how many zeros will it end?
Page 1 of 4
Continued on the next page...
Math 147
Exercises (continued)
Solution: We need to compute r such that 10r k2010!. But 10 is not prime so we cannot
just copy the approach in the previous discussion. Note that 10 is formed by a 2 and a 5.
So we need to count how many pairs of 2 and 5 can be found in the prime factorization of
2010!.
Since, for any n, b 2ni c ≥ b 5ni c, it suffices to compute for r such that 5r k2010!. Hence, the
number we are looking for is Hence, the number we are looking for is
∞
X
2010
b i c
5
i=1
= b
=
2010
2010
2010
2010
2010
2010
2010
c+b
c+b
c+b
c+b
c + b 6 c + b 7 c + ···
5
25
125
625
3125
5
5
501.
2. Let n be a positive integer.
(a) Show that the binomial coefficient
cn =
2n
n
is even.
(b) Prove that cn is divisible by 4 if and only if n is not a power of 2.
r
Solution:
2s k(n!). It suffices to show that r > 2s. From the previous examples,
P∞ Let 2 k(2n)! and
P∞
n
r = i=1 b 2n
c
while
s
=
i=1 b 2i c.
2i
Observe that if n = 2i q + a, 0 ≤ a < 2i , we have
a
n
= q + i,
2i
2
0≤
a
< 1.
2i
Hence, q = b 2ni c. Likewise
a
2n
= 2q + i−1 ,
i
2
2
where
(
0≤
1≤
a
2i−1
a
2i−1
< 1, if a < 2i−1 ,
< 2, if 2i−1 ≤ a < 2i .
Therefore,
2n
b ic=
2
(
2b 2ni c,
if a < 2i−1 ,
n
2b 2i c + 1, if 2i−1 ≤ a < 2i .
n
a. Since b 2n
2i c ≥ 2b 2i c, it suffices to show that there exist an i for which the inequality strictly
hold.
Suppose that 2b ≤ n < 2b+1 . Putting i = b + 1 and applying the result above, imply that
n
2n
c = 2b b+1 c + 1 = 1,
2b+1
2
n
Since b 2b+1
c = 0, we have r ≥ 2s and 2n
n is even.
b
b. We need to show that for n not a power of 2, r ≥ 2s + 2. From (a.) we have already seen
that
2n
n
b b+1 c = 2b b+1 c + 1 = 1.
2
2
It suffices to find an i 6= b + 1, such that
b
Page 2 of 4
2n
n
c = 2b i c + 1.
2i
2
Continued on the next page...
Math 147
Exercises (continued)
Suppose that 2b < n < 2b+1 and assume n = 2b + m where 0 ≤ m < 2b . Choose c such
that 2c ≤ m < 2c+1 . Note that 0 ≤ c < c + 1 ≤ b. Applying the above result to m and to
c + 1, we get
m
2m
b c+1 c = 2b c+1 c + 1 = 1.
2
2
Therefore,
n
2c+1
=
2b
2c+1
+
m
2c+1
m
n
= 2b−c−1 + c+1
2c+1
2
n
b c+1 c = 2b−c−1 .
2
n
2b c+1 c = 2b−c .
2
and
2b+1
2m
2n
=
+ c+1
2c+1
2c+1
2
2n
b c+1 c = 2b−c + 1.
2
n
2n
b c+1 c = 2b c+1 c + 1.
2
2
3. (a) Find the number of ordered triple (a, b, c) of positive integers such that lcm(a, b) = 1000,
lcm(b, c) = 2000 and lcm(a, c) = 2000.
(b) For three nonzero integers a, b and c, show that
(a, [b, c]) = [(a, b), (a, c)].
(c) Prove that
lcm(a, b, c)2
gcd(a, b, c)2
=
lcm(a, b) lcm(b, c) lcm(c, a)
gcd(a, b) gcd(b, c) gcd(c, a)
Solution:
a. Since a|[a, b], every prime p that divide a, also divides [a, b]. We can apply the same
reasoning to b and c and see that a = 2r1 5s1 , b = 2r2 5s2 and c = 2r3 5s3 for some nonnegative integers r1 , r2 , r3 , s1 , s2 , and s3 . Then [a, b] = 2max(r1 ,r2 ) 5max(s1 ,s2 ) , [a, c] =
2max(r1 ,r3 ) 5max(s1 ,s3 ) and [b, c] = 2max(r2 ,r3 ) 5max(s2 ,s3 ) . Therefore,
max(r1 , r2 ) = 3
max(r1 , r3 ) = 4
max(r2 , r3 ) = 4
max(s1 , s2 ) = 3
max(s1 , s3 ) = 3
max(s2 , s3 ) = 3
Therefore, the possibilities are
r1
r2
r3
Page 3 of 4
0
3
4
1
3
4
2
3
4
3
0
4
3
1
4
3
2
4
3
3
4
Continued on the next page...
Math 147
Exercises (continued)
and
s1
s2
s3
0
3
3
1
3
3
2
3
3
3
0
3
3
1
3
3
2
3
3
3
0
3
3
1
3
3
2
3
3
3
Therefore, we have 7* 10 =70 possible solutions.
b. We demonstrate here a technique for proving equality of positive integers. Note that for
positive integers a and b, a|b and b|a if and only if a = b. Also, a|b means that for every
prime p such that pr ka, we hve pr |b.
Suppose that p is a prime such that pr k(a, [b, c]). Then pr |a and pr |[b, c]. The latter means
that pr |b or pr |c. Therefore, pr |a and (pr |b or pr |c). Hence, (pr |a and pr |b) or (pr |a and
pr |c). This implies that pr |(a, b) or pr |(a, c), which in turn implies that pr |[(a, b), (a, c)].
Therefore, (a, [b, c])|[(a, b), (a, c)].
Suppose that p is a prime such that pr k[(a, b), (a, c)]. Then pr |(a, b) or pr |(a, c). Which in
turn means that (pr |a and pr |b) or (pr |a and pr |c). The latter means that pr |a and (pr |b
or pr |c). That is, pr |a and pr |[b, c], or equivalently pr |(a, [b, c]). Thus, (a, [b, c]) divides
[(a, b), (a, c)] and equality follows,
Q
Q
Q
c. Let a = p prime plp , b = p prime pmp and c = p prime pnp . Then
[a, b, c] =
Y
pmax(lp ,mp ,np )
p a prime
[a, b] =
Y
pmax(lp ,mp )
p a prime
[a, c] =
Y
pmax(lp ,np )
p a prime
[b, c] =
Y
pmax(mp ,np )
p a prime
(a, b, c) =
Y
pmin(lp ,mp ,np )
p a prime
(a, b) =
Y
pmin(lp ,mp )
p a prime
(a, c) =
Y
pmin(lp ,np )
p a prime
(b, c) =
Y
pmin(mp ,np )
p a prime
Therefore,
Q
2 max(lp ,mp ,np )
[a, b, c]2
p a prime p
=Q
max(a,b)+max(a,c)+max(b,c)
[a, b][a, c][b, c]
p a prime p
Y
=
p2 max(lp ,mp ,np )−max(a,b)−max(a,c)−max(b,c)
p a prime
Q
2 min(lp ,mp ,np )
(a, b, c)
p a prime p
=Q
min(a,b)+min(a,c)+min(b,c)
(a, b)(a, c)(b, c)
p a prime p
Y
=
p2 min(lp ,mp ,np )−min(a,b)−min(a,c)−min(b,c)
2
p a prime
Page 4 of 4
December 11, 2010\jbasilla\201010math117.tex
End of Examination
Math 147
Exercises (continued)
It remains to show then that for any three nonnegative numbers p, q, r
2 max(p, q, r)−max(p, q)−max(p, r)−max(q, r) = 2 min(p, q, r)−min(p, q)−min(p, r)−min(q, r).
Without loss of generality, we can assume p ≤ q ≤ r. Equip with this assumption, we
obtain
2 max(p, q, r) − max(p, q) − max(p, r) − max(q, r) = 2r − q − r − r = −q
2 min(p, q, r) − min(p, q) − min(p, r) − min(q, r) = 2p − p − p − q = −q.
This proves the equality.
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Continued on the next page...