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Texas A&M High School Math Contest
Best Student Exam Solutions 2008
1. Let f (x) = |x − 1| + |x − 2| + · · · + |x − 2008|. Find the set of all real numbers x
such that f (x) is a minimum.
Solution: Note that f (x) is convex and piece-wise linear. Then the minimum
is achieved at the segment where f (x) is constant. This is possible only if 1004 ≤
x ≤ 1005 and therefore the set of all real numbers x such that f (x) = min f (y) is the
y
closed interval [1004, 1005].
π 2
2. Find value of 12 − sin2 16
.
Solution: We have
Therefore
− sin2
π
16
=
1
π
− sin2
2
16
2
1
2
1
2
1−cos
2
−
√
11+
=
4 2
π
8
2
2
=
1
2
cos π8 and cos2
π
8
=
1+cos
2
π
4
.
√
2+ 2
=
.
16
3. Find the last three digits of 20072008 .
Solution: We first note that 20072008 ≡ 72008 (mod 1000). Then we use that
φ(1000) = 400, where φ(n) is Euler’s totient function. According to Euler’s theorem,
we get 7400 ≡ 1 (mod 1000). Then, we obtain 20072008 ≡ 78 (mod 1000). A simple
computation gives
78 = (50 − 1)4 = 504 − 4 (50)3 + 6 (50)2 − 200 + 1.
Then 78 ≡ −199 ≡ 801 (mod 1000) and the last three digits of 20072008 are 801.
4. Solve for x:
√
p
p
√
x2 − (x − 1)2 + (x − 2)2 = 5.
√
Solution: The equation simplifies to |x| − |x − 1| + |x − 2| = 5. There are only
two solutions:
√
√
(i) if x ≤ 0 then we solve −x − (1 − x) + 2 − x =√ 5 and obtain x = 1 −√ 5,
(ii) if x ≥ 2 then we solve x −√(x − 1) + x − 2 = 5 and obtain x = 1 + 5,
because the left side is less than 5 in the interval 0 < x < 2:
√
|x| − |x − 1| + |x − 2| ≤ 2 < 5.
5. Let p, q, r, s, t be the numbers 1, 2, 3, 4, 5, but not necessarily in that order. Set
x=
1
p+
.
1
q+
1
r+
1
s+ 1
t
What choice of t makes x as large as possible?
Solution: To make x as large as possible, make the denominator as small as possible. Since the denominator is p plus a fraction y less than 1, we must have p = 1. To
make y as large as possible, note that since it has denominator q plus a fraction z less
than 1, we must take q as large as possible. Hence q = 5. We also need z as large as
possible, so we must take r = 2. Continuing in this way, we must take s = 4 and t = 3.
6. Let
2
999
1
+ + ··· +
.
2! 3!
1000!
Which of the following inequalities is true?
x=
a) x < 0.999
b) 0.999 ≤ x < 1 − 10−12345
c)1 − 10−12345 ≤ x < 1.0
d) 1.0 ≤ x < 1 + 10−12345
Solution: Observe that
x=
2−1 3−1
1000 − 1
1
1
1
1
1
1
+
+ ··· +
= − + − + ··· +
−
.
2!
3!
1000!
1! 2! 2! 3!
999! 1000!
This is a telescoping sum and simplifies to x = 1 − 1/1000!. Since 1000! is a product
of 1000 numbers, each of which is at most 1000, we have
1000! ≤ 10001000 = 103000 < 1012345 .
Thus 0.999 < x < 1 − 10−12345 and the correct answer is b).
7. Find the radius of the largest ball that can be fit inside
B = {(x, y, z)| x2 + y 2 + z 2 ≤ 1, x ≥ 0, y ≥ 0, z ≥ 0}.
Solution: Let O = (0, 0, 0), A be the center of that ball and r be its radius. It
is clear that the ball must touch all coordinate planes and the unit sphere. Let B be
the point where the ball touches the unit sphere. Then the points O, A, and B are
on the same line and OA + AB = 1. We know that AB = r and because the ball
touches all coordinate planes the segment OA is the main diagonal in a cube with
side length r. Therefore OA2 = 3r2 and we find r = 1+1√3
8. For how many n in the set {1, 2, 3, . . . , 100} is the tens digit of n2 odd?
2
Solution: There are two in each decile, namely 10a+4 and 10a+6, giving a total
of twenty. This is because the tens digit of (10a + 4)2 = 100a2 + 80a + 16 is the units
digit of 8a + 1, while the tens digit of (10a + 6)2 = 100a2 + 120a + 36 is the units
digit of 2a + 3, both of which are odd for any integer a. All the other tens digits of
perfect squares are even: (10a + b)2 = 100a2 + 20a + b2 , the tens digit of which is the
tens digit of 2a + b2 , which is even if the tens digit of b2 is even. But the tens digit
of b2 is even if b 6= 4, 6.
9. Find the value of
∞
X
2i + 1
.
2
2
i
(i
+
1)
i=1
Solution: Using that
1
1
2i + 1
= 2−
2
+ 1)
i
(i + 1)2
i2 (i
we obtain
∞
∞
∞
X
X
2i + 1
1 X1
=
−
= 1.
i2 (i + 1)2
i2
i2
i=1
i=1
i=2
10. If logy x + logx y = 8, then find the value of (logy x)2 + (logx y)2 .
Solution: Let z = logy x. Then the problem reduces to computing the value of
1
1
z 2 + 2 given that z + = 8. Then, we derive
z
z
2
1
1
2
z + 2 = z+
− 2 = 82 − 2 = 62.
z
z
11. Find the limit
S = lim n−2009 1 + 22008 + 32008 · · · + n2008 = lim Sn
n→∞
n→∞
Solution: We note that
2008
n
X
1 i
Sn =
n n
i=1
is a Riemann sum for the integral
R1
0
x2008 dx. Hence
Z
lim Sn =
n→∞
1
x2008 dx =
0
3
1
.
2009
12. Let a and b be real numbers such that a + b = 1 and (1 − 2ab)(a3 + b3 ) = 12.
What is the value of a2 + b2 ?
Solution: Let x = a2 + b2 . the we have 1 − 2ab = (a + b)2 − 2ab = x and
a3 + b3 = (a + b)(a2 − ab + b2 ) = x − ab = x −
1−x
3x − 1
=
.
2
2
Then the equation (1 − 2ab)(a3 + b3 ) = 12 reduces to x(3x − 1) = 24. The two
solutions are x = 3 and x = − 38 . Using that x = a2 + b2 ≥ 0, we conclude a2 + b2 = 3.
13. If 2008k divides 2007! and 2008k+1 does not, then what is the value of k?
Solution: We have that 2008 = 23 · 251 and 251 is a prime number. Then the
power k is equal to the integer part of 2007
because 2512 does not divide any of the
251
numbers 1, 2 . . . 2007. That gives k = 7.
14. Adriel has a collection of blue marbles and green marbles. The number of blue
marbles is either 2, 5, 6, 7, 11, or 13. If two marbles are chosen simultaneously and at
random from his collection, then the probability they have different color is 21 . How
many blue marbles are in Adriel’s collection?
Solution: Let b be the number of blue marbles
and g be the number of the green
marbles. We can choose two marbles in b+g
ways.
Hence, we have
2
bg
2bg
1
= b+g =
2
(b + g)(b + g − 1)
2
We write this as a quadratic equation for g
g 2 − (2b + 1)g + b2 − b = 0.
The above equation must have integer roots. Therefore the discriminant D, which
should be an integer, is also a perfect square, say n2 . That is
D = (2b + 1)2 − 4(b2 − b) = 8b + 1 = n2 .
The only value of blue marbles that gives a perfect square is b = 6.
15. In ∆ABC, the side lengths are AB = 2008, BC = 1004, and CA = 1506. Side
BC is extended to point D so that ∆DAB is similar to ∆DCA. What is DC?
Solution: Using that the triangles ∆DAB and ∆DCA are similar we derive
AD
AB
BD
=
=
.
CD
AC
AD
4
Note that if x = DC, then BD = 1004 + x, and we get
AB
2008x
4x
AD = x
=
=
,
AC
1506
3
1506
3(1004 + x)
AC
= (1004 + x)
=
.
AD = BD
AB
2008
4
From the above identities we derive the equation
4x
3(1004 + x)
=
.
3
4
9036
This gives x =
.
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16. Let f (x) be a function such that f (x) + 3f (−x) = cos x for every real number x.
What is the value of f (π)?
Solution: We use x = π and obtain f (π) + 3f (−π) = cos π = −1. Similarly, we
get f (−π) + 3f (π) = −1 when x = −π. Eliminating f (−π) from the two equations,
we obtain f (π) − 9f (π) = −1 + 3 = 2. Hence, we have f (π) = − 41 .
17. Calculate the integral
Z
1
ex sin(πx) dx.
0
Solution: Integrating by parts we get
Z
Z 1
x
e sin(πx) dx = −π
I=
1
ex cos(πx) dx = −πJ.
0
0
We integrate by parts again and obtain
Z 1
x
1
J = e cos(πx)|0 + π
ex sin(πx) dx = −e − 1 + πI.
0
Therefore, I satisfies
I = −π (−e − 1 + πI)
and we derive I =
π(e + 1)
π2 + 1
18. How many triples (x, y, z) satisfy the equations
xy = 2z, xz = 2y, and yz = 2x?
Solution: We first observe that if one of the unknowns is zero then all of them
are zero. Hence, (0, 0, 0) is one triple. Assume now that x, y and z are all non-zero.
Then x2 yz = xy · xz = 2z · 2y = 4yz and we get x2 = 4 because yz is not zero.
Similarly, we have y 2 = 4 and z 2 = 4. If x = 2 we have y = z = 2 or y = z = −2 and
in the other case, x = −2 we have y = −z = 2 or y = −z = −2. Therefore, we have
five different triples: (0, 0, 0), (2, 2, 2), (2, −2, −2), (−2, 2, −2), (−2, −2, 2).
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