Download Best Student Solutions

Texas A&M High School Math Contest
Best Student Exam Solutions 2007
1. Solve the equation |x − 1| + |x − 2| + |x − 3| = 4.
}.
Solution: There are two values for x which satisfy the above equation: { 32 , 10
3
One can get these by solving 1 − x + 2 − x + 3 − x = 4 and x − 1 + x − 2 + x − 3 = 4.
Because f (x) = |x − 1| + |x − 2| + |x − 3| − 4 is a convex function it can intersect the
x-axis at most two times.
2. Let tan(α) =
90 degrees.
√
6+
√
3−
√
2 − 2. Find the angle α if it is an angle between 0 and
Solution: It is not hard to check by hand that the angle is between 30 and 45
degrees. Then the first angle to try is 37.5 and it works. For example, we take θ = 75
is the formula
θ
1 − cos θ
tan =
2
sin θ
√
√
√
and simplify the result to get tan(37.5) = 6 + 3 − 2 − 2.
3. Find the smallest number N such that its decimal representation is N = aa . . . a
and it is divisible by 29.
Solution: First a = 1 because 29 is a prime. Then, we use that 1028 gives
remainder 1 when divided by 29. The rest is to check that for any power k less than
28 the remainder of 10k (mod 29) is not 1. Thus N consists of a string of 28 ones and
must equal (1028 − 1)/9. We need to verify that property for only divisors of 28. It
is easy to check that:
100 has remainder 13 when divided by 29. We will write 100 ≡ 13 (mod 29)
100 ≡ 13 (mod 29)
1000 ≡ 130 ≡ 14 (mod 29)
104 ≡ 140 ≡ 130 + 10 ≡ 14 + 10 = 24 (mod 29)
105 ≡ 240 ≡ 100 + 140 ≡ 13 + 24 ≡ 37 ≡ 8 (mod 29)
107 = 100 · 105 ≡ 104 ≡ 17 (mod 29)
1014 ≡ 172 ≡ −1 (mod 29)
4. Solve for x:
4x − 3x−1/2 = 3x+1/2 − 22x−1
C
D
1
a
O
A
B
a 2
2
Solution: We rewrite the equation in the following way
√ 2x−1 √ 2x+1 4x
x
4 −
3
3
=
−
2
and obtain
3
Therefore
3x
4x
= 2√ .
2
3
x 23
4
4
=
3
3
and this gives x = 1.5.
5. Find the last two digits of 7555 .
Solution: We start with 752 ≡ 25 (mod 100) and 252 ≡ 25 (mod 100). Therefore,
we have 7554 ≡ 2527 ≡ 25 (mod 100), and we obtain 7555 ≡ 75 (mod 100).
6. Solution: The limit is limx→0 xx = limx→0 ex ln x = 1 because
lim x ln x = lim
x→0
x→0
ln x
1
x
= lim
x→0
1
x
− x12
= 0.
7. Find the volume of the largest cube with sides parallel to the coordinate planes
that can be fit inside B.
B = {(x, y, z)| x2 + y 2 + z 2 ≤ 1, z ≥ 0}
Solution: It is easy to see that the largest cube of this type must have its base
on the xy-plane and the center of the base (the intersection of the diagonals) is the
2
point (0, 0, 0). Moreover, the top side of the cube must touch the sphere B. If this is
not the case then the cube could be enlarged. Let ABCD be a rectangle such that:
(i) AB is a diagonal in the base of the cube; (ii) AD, DC and CB are edges of the
cube, see Figure
√ 1. Let’s denote the length of the edge of the cube with a. We have
that |AB| = a 2, |AD| = |DC| = |CB| = a, AB lies on a diameter of the unit circle,
and D and C are on the circumference, see Fig. .
Let O be the center of the circle (note that |AO| = |OB|). Then, the triangle
AOD is a right triangle with sides
√
a 2
, |OD| = 1, |AD| = a.
|AO| =
2
Using the equation |AO|2 + |AD|2 = |OD|2 we derive
3a2
=1
2
which gives a =
q
2
.
3
Therefore, the volume V of the cube is
√
32
2
2 2
V =a =
= √
3
3 3
3
8. Fix 0 < x0 < 1. A sequence is defined by xi+1 =
is the fractional part of a. For example
1
xi
for i = 0, 1, . . . , where {a}
8
1
{1.3} = 0.3 and
=
7
7
Find the largest possible x0 such that x0 = x1 = x2007 .
Solution: Let x10 = a + x0 where a is a positive integer. Then
have the following equation for x0
x20 + ax0 − 1 = 0.
Using that 0 < x0 < 1, we obtain
√
2
2
−a + a2 + 4
√
√ .
≤
=
x0 =
2
2
a+ a +4
1+ 5
Hence,
2
√ =
x0 =
1+ 5
3
√
5−1
2
n o
1
x0
= x0 and we
and for that initial value, we have x0 = xk for any k ≥ 1.
9. Find the value of
∞
X
1
i(i + 1)(i + 2)
i=1
P
Solution: This is a convergent series because ∞
i=1
partial fractions, we get the representation
1
i(i+1)(i+2)
≤
P∞
1
i=1 i3 .
Using
1
1
1
1
=
−
+
.
i(i + 1)(i + 2)
2i i + 1 2(i + 2)
We substitute the above in the sum
∞
X
1
1 1
1
1 1 1
1 1 1
+
+··· ,
=
·1−1· + ·
· −1· + ·
i(i + 1)(i + 2)
2
2 2 3
2 2
3 2 4
i=1
and rearrange the summation to conclude
∞
X
i=1
1 1
1
1
= − = .
i(i + 1)(i + 2)
2 4
4
10. Given the points A(1, 32 ) and B(4, 5) find the minimum distance |AC| + |BC|
where C is a point on the line 2x + y = 1.
Solution: Let A′ be the reflection of A about the line 2x + y = 1. It is easy to
verify that A′ (−1, 21 ). Given any point C on the line 2x + y = 1, we have
|AC| + |BC| = |A′ C| + |BC| ≥ |A′ B|
because A′ BC is a triangle and the sum of any two sides in a triangle is greater or
equal to the third side. Note that equality is possible only when C is the intersection
point of the segment A′ B and the line 2x + y = 1. In any case, we have that |A′ B| is
the minimum distance and
r
√
181
81
′
|A B| = 25 +
=
.
4
2
11. Find the limit
1 + 1/2 + · · · + 1/n
n→∞
log(n)
Solution: For n ≥ 2, we have that
Z n
1
1
1
1
1
<
dx = ln(n) < 1 + + · · · + .
1+ +···+
2
n−1
2
n
1 x
lim
4
Then
1≤
ln(n + 1)
1 + 1/2 + · · · + 1/n
≤
ln(n)
ln(n)
and we derive
lim
n→∞
Therefore
1 + 1/2 + · · · + 1/n
= 1.
ln(n)
1 + 1/2 + · · · + 1/n
= ln 10
n→∞
log(n)
lim
12. Solve the equation
2e2|x|−1 + 2e1−2|x| = 7 − 12x2
Solution: Both sides are even function of x. Therefore, if x solves the equation
then −x also solves it. Hence, we will consider only the case x ≥ 0. Let f (x) =
2e2|x|−1 + 2e1−2|x| and g(x) = 7 − 12x2 . Using that x ≥ 0 and 2 < e < 3, we have that
f (x) = 2e2x−1 + 2e1−2x ,
f (0) =
2
+ 2e < 6 + 1 = 7 = g(0).
e
The function f is concave up on [0, ∞) (compute f ′ and f ′′ ) with a global minimum at
x = 21 : f ( 21 ) = 4. On the other hand, the function g is concave down and decreasing
on [0, ∞) with g( 21 ) = 4. Based on the properties:
(i) f (0) < g(0),
(ii) f is concave up and g is concave down,
(iii) g on decreasing [0, ∞) with g( 21 ) = f ( 12 ) = 4
we conclude that the graphs of f and g can intersect only once for x ≥ 0 and the
intersection point is ( 21 , 4). Therefore, the only solutions of the equation are {− 21 , 12 }
13. We flip a regular coin 2007 times and record the outcome: heads or tails. The
1
probability of getting heads or tails each time is . Let p1 be the probability that
2
the number of heads recorded at the end is twice the number of tails and p2 be the
p1
probability that the number of tails recorded at the end is 1338. Find the ratio .
p2
Solution: Note that ”the number of heads recorded at the end is twice the number
of tails” gives that the number of tails is 669. Therefore
1
1
2007
2007
p1 = 2007
= 2007
= p2
2
669
2
1338
and we conclude
p1
= 1.
p2
14. Find the number of zeros at the end of 2007!
5
Solution: We need to find the power
of 5 that divides 2007!. The way we count
2007
this is the following: we have 5 numbers less than 2007 divisible by 5, we also
have 2007
numbers less than 2007 divisible by 25, and so on... Note that the number
52
2007
is less than one. Therefore, we derive that power of 5 that divides 2007! is
55
∞ X
2007
k=1
5k
= 401 + 80 + 16 + 3 = 500.
Similarly, we can find the power of 2 that divides 2007! but that number is clearly
bigger. Therefore, the number of zeros at the end of 2007! is 500.
15. Find the smallest of the following numbers
2008
20062007
2008
, 20072006
2006
, 20072008
2007
, 20062008
Solution: We will use the inequality
ab < ba
for any integers 2 < b < a. We first observe
2008
> 20062008
2008
> 20072008
20062007
and
20072006
2007
2006
.
20082006
In order to show that the smallest number is 2007
2007
20062008
2006
> 20072008
, we need to prove
.
Taking ln of both sided, we get the equivalent inequality
20082007 ln(2006) > 20082006 ln(2007)
which holds because
2008 ln(2006) > 2 ln(2006) = ln(20062 ) > ln(2007).
2006
Therefore, the smallest integer is 20072008
.
16. Find the smallest integer larger than
√
√
√
√
√
√
S = 1024 − 1025 + 1026 − 1027 + · · · − 2025 + 2026
6
Solution: We rearrange
S=
Note that
√
√
2026 −
1012
X
√
k=512
2k + 1 −
√
2k =
√
2026 − δ.
1024 = 32 and
44 <
√
2026 = 45 +
√
2026 −
√
2025 = 45 + √
1
√
< 45.02
2026 + 2025
We also have
√
√
1
1
1
< n+1− n= √
√ < √ .
2 n
n+1+ n
2 n+2
√
Therefore δ can be bounded by
δ1 :=
1012
1012
1012
X
√
√
1 X 1
1
1 X
√
√ =: δ2 .
2k + 1 − 2k ≤
≤
2 k=512 2k + 2 k=512
2 k=512 2k
The difference between δ1 and δ2 is small:
√
1
2026 − 32
1
1
√
√
δ2 − δ1 =
=
−√
.
2
1024
2026
64 2026
√
Using the bounds for 2026, we get
δ2 − δ1 <
45.02 − 32
1
<
.
64 · 45
100
we observe that 4δ1 and 4δ2 are lower and upper Darby sums for the integral
RNow,
2026 √
x dx. Therefore
1024
1
δ1 <
4
Z
2026
1024
1
√ dx =
x
√
2026 −
2
√
1024
< δ2 .
The sum S can be approximated by
√
√
√
√
√
2026 − 1024
2026 + 1024
∗
=
S = 2026 −
2
2
with the following error estimate
|S − S ∗ | < δ2 − δ1 <
7
1
= 0.01.
100
Using the bounds for
√
2026, we get
√
√
45.02 + 32
2026 + 1024
45 + 32
∗
<S =
<
= 38.51
38.5 =
2
2
2
Therefore, the real sum S can be estimated by
38.49 < S < 38.52
and the smallest integer bigger than S is 39.
17. Calculate the integral
Z
0
1
207 200
x (1 − x)7 dx
7
Solution: Integration by parts gives
Z 1 Z 1 207 201
207 200
7
x (1 − x)6 dx
x (1 − x) dx =
6
7
0
0
and repeating the integration by parts until (1 − x) disappears we get
Z 1
Z 1 1
207 200
7
x (1 − x) dx =
x207 dx =
7
208
0
0
18. Calculate the sum of the squares of the real roots of
x2007 − x2005 − x2003 − · · · − x3 − 2x = 0
Solution: One real root is x = 0 and the rest of the real roots solve the reduced
equation
x2006 − 1 = x2004 + x2002 + · · · + x2 + 1.
Using that |x| = 1 does not solve the equation, we sum the finite geometric series in
the left side to obtain
x2006 − 1
x2006 − 1 = 2
.
x −1
Rearranging the above equation, we get
x2 − 2
x2006 − 1
= 0.
x2 − 1
x2006 − 1
does not have any real zeros. Therefore, we conclude that the
x2 − 1
√ √
only real zeros of the original equation are {0, − 2, 2} and the sum of their squares
is 4.
The term
8