Problem 1. If x, y, and z are positive integers such that xy = 20 and yz = 12, compute the smallest possible value of x + z. Solution 1. Note that x and z can each be minimized by making y as large as possible, so set y = lcm(12, 20) = 4. Then
x = 5, z = 3, and x + z = 8.
Problem 2. Let T = T N Y W R. Let A = (1, 5) and B = (T − 1, 17). Compute the value of x such that (x, 3) lies on the perpendicular bisector of AB.
Solution 2. The midpoint of AB is (T2, 11), and the slope of ←→AB is T −212 . Thus the perpendicular bisector of AB has slope 2−T
12 and passes through the point ( T
2, 11). Thus the equation of the perpendicular bisector of AB is
y = (2−T12 )x + (11 − 2T −T24 2). Plugging y = 3 into this equation and solving for x yields x = T −296 +T2. With T = 8, it follows that x = 966 +82 = 16 + 4 = 20.
Problem 3. Let T = T N Y W R. Let N be the smallest positive T -digit number that is divisible by 33. Compute the product of the last two digits of N .
Solution 3. The sum of the digits of N must be a multiple of 3, and the alternating sum of the digits must be a multiple of 11. Because the number of digits of N is fixed, the minimum N will have the alternating sum of its digits equal to 0, and therefore the sum of the digits of N will be even, so it must be 6. Thus if T is even, then N = 10 . . . 0
| {z }
T −3 0’s
23, and if T is odd, then N = 10 . . . 0 | {z }
T −3 0’s
32. Either way, the product of the last two digits of N is 6 (independent of T ).
Problem 4. Let T = T N Y W R. In the square DEF G diagrammed at right, points M and N trisect F G, points A and B are the midpoints of EF and DG, respectively, and EM ∩ AB = S and DN ∩ AB = H. If the side length of square DEF G is T , compute [DESH].
S H N M B A G F D E
Solution 4. Note that DESH is a trapezoid with height T2. Because AS and BH are midlines of triangles EF M and DGN respectively, it follows that AS = BH = T6. Thus SH = T − 2 ·T6 =2T3 . Thus [DESH] = 21(T +2T3 ) ·T2 = 5T122. With T = 6, the desired area is 15.
Problem 5. Let T = T N Y W R. For complex z, define the function f1(z) = z, and for n > 1, fn(z) = fn−1(z). If
f1(z) + 2f2(z) + 3f3(z) + 4f4(z) + 5f5(z) = T + T i, compute |z|.
Solution 5. Because z = z, it follows that fn(z) = z when n is odd, and fn(z) = z when n is even. Taking z = a + bi, where
a and b are real, it follows that P5
k=1kfk(z) = 15a + 3bi. Thus a = 15T, b = T3, and |z| =
√
a2+ b2= |T | √
26 15 .
With T = 15, the answer is√26.
Problem 6. Let T = T N Y W R. Compute the number of ordered pairs of positive integers (a, b) with the property that ab = T20· 21012, and the greatest common divisor of a and b is 1.
Solution 6. If the prime factorization of ab is pe1
1 p e2
2 . . . p ek
k , where the pi’s are distinct primes and the ei’s are positive
integers, then in order for gcd(a, b) to equal 1, each pi must be a divisor of exactly one of a or b. Thus the
desired number of ordered pairs is 2k because there are 2 choices for each prime divisor (i.e., pi | a or pi | b).
With T =√26, it follows that (√26)20· 21012= (210· 1310) · 21012= 222· 312· 512· 712· 1310. Thus there are
five distinct prime divisors, and the answer is 25= 32. Problem 7. Let T = T N Y W R. Given that sin θ =
√ T2−64
T , compute the largest possible value of the infinite series
Solution 7. Using sin2θ + cos2θ = 1 gives cos2θ = T642, so to maximize the sum, take cos θ =
8
|T |. Using the formula for
the sum of an infinite geometric series gives 1−8/|T |8/|T | =|T |−88 . With T = 32, the answer is 248 = 13.
Problem 15. Given that April 1st, 2012 fell on a Sunday, what is the next year in which April 1st will fall on a Sunday?
Solution 15. Note that 365 = 7 · 52 + 1. Thus over the next few years after 2012, the day of the week for April 1st will
advance by one day in a non-leap year, and it will advance by two days in a leap year. Thus in six years, the day of the week will have rotated a complete cycle, and the answer is 2018.
Problem 14. Let T = T N Y W R, and append the digits of T to A A B (for example, if T = 17, then the result would be 1 7 A A B). If the resulting number is divisible by 11, compute the largest possible value of A + B.
Solution 14. Let R be the remainder when T is divided by 11. Note that the alternating sum of the digits of the number must be divisible by 11. This sum will be congruent mod 11 to B − A + A − R = B − R, thus B = R. Because A’s value is irrelevant, to maximize A + B, set A = 9 to yield A + B = 9 + R. For T = 2018, R = 5, and the answer is 9 + 5 = 14.
Problem 13. Let T = T N Y W R, and let K = T − 2. If K workers can produce 9 widgets in 1 hour, compute the number of workers needed to produce 720K widgets in 4 hours.
Solution 13. Because T workers produce 9 widgets in 1 hour, 1 worker will produce T9 widgets in 1 hour. Thus 1 worker will produce 36T widgets in 4 hours. In order to produce 720T widgets in 4 hours, it will require 720/T36/T = 20 workers (independent of T ).
Problem 12. Let T = T N Y W R. For some real constants a and b, the solution sets of the equations x2+(5b−T −a)x = T +1
and 2x2+ (T + 8a − 2)x = −10b are the same. Compute a.
Solution 12. Divide each side of the second equation by 2 and equate coefficients to obtain 5b − T − a = T2 + 4a − 1 and T + 1 = −5b. Thus b = T +1
−5 , and plugging this value into the first equation yields a = − T
2. With T = 20, the
answer is −10.
Problem 11. Let T = T N Y W R. Given that log248!+ log
428!= 6! · T · x, compute x.
Solution 11. Note that 48! = 22·8!, thus log
248! = 2 · 8!. Similarly, log428!= 8!2. Thus 2 · 8! + 8!
2 = 6!(2 · 7 · 8 + 7 · 8
2) = 6! · 140.
Thus 140 = T x, and with T = −10, x = −14.
Problem 10. Let T = T N Y W R, and let d = |T |. A person whose birthday falls between July 23 and August 22 inclusive is called a Leo. A person born in July is randomly selected, and it is given that her birthday is before the dth
day of July. Another person who was also born in July is randomly selected, and it is given that his birthday is after the dth day of July. Compute the probability that exactly one of these people is a Leo.
Solution 10. Note that there are 9 days in July in which a person could be a Leo (July 23–31). Let the woman (born before the dth day of July) be called Carol, and let the man (born after the dth day of July) be called John, and
consider the possible values of d. If d ≤ 21, then Carol will not be a Leo, and the probability that John is a Leo is 31−d9 . If d = 22 or 23, then the probability is 1. If d ≥ 24, then John will be a Leo, and Carol will not be a Leo with probability 1 − d−23d−1. With T = −14, the first case applies, and the desired probability is 179. Problem 9. Let T = T N Y W R. When T is expressed as a reduced fraction, let m and n be the numerator and denominator,
respectively. A square pyramid has base ABCD, the distance from vertex P to the base is n − m, and P A = P B = P C = P D = n. Compute the area of square ABCD.
Solution 9. By the Pythagorean Theorem, half the diagonal of the square is pn2− (n − m)2 = √2mn − m2. Thus the
diagonal of the square is 2√2mn − m2, and the square’s area is 4mn − 2m2. With T = 9
17, m = 9, n = 17, and
Problem 8. Let R be the larger number you will receive, and let r be the smaller number you will receive. In the diagram at right (not drawn to scale), circle D has radius R, circle K has radius r, and circles D and K are tangent at C. Line←→Y P is tangent to circles D and K. Compute Y P .
P Y
D C K
Solution 8. Note that DY and KP are both perpendicular to line ←→Y P . Let J be the foot of the perpendicular from K to DY . Then P KJ Y is a rectangle and Y P = J K =√DK2− DJ2 =p(R + r)2− (R − r)2= 2√Rr. With