Mock AIME 3 2010 Problems

Mock AIME III By: Zhero Brut3Forc3 andersonw dysfunctionalequations April 1, 2010 Negative AIME 1 -0. Find the value of 645. -1. Find the sum of all distinct integers n such that n|243. -2. Find the number of ways a teacher can distribute 1 piece of candy among 243 hungry schoolchildren sitting in a row such that no child receives more than one piece of candy, and that no two children that sit next to each other may both receive candy. -3. We have a triangle abc with sides ab, bc, ca with lengths α, β, and γ, respectively, so that α + β + γ = π. The measures of angles ∠bac, ∠acb, and ∠cba are A, B, and C, respectively, and the perimeter of the triangle with sides AB, BC, and CA is π 2 3 . Let ω, with center I, be the circumcircle of 4abc, and let Γ, with center O, be the incircle of 4abc. Let [ be the circle tangent to Γ, ω, and bc, let ] be the circle tangent to Γ, ω, and ca, and let \ be the circle tangent to Γ, ω, and ac. Let [, ], and \ be tangent to ω at , ×, and , respectively. Find the largest integer less than or equal to the largest possible value of the area of 4

×.

-4. Let f(x, y, z) = xyz + 49(x + y + z) + 1337 xy + yz + zx + 240 . Let g(1) = f(1, 1, 1), and let g(n) = f(g(n − 1), n, n) for n ≥ 2. Find g(2010). -5. Let f(x) = x 2 . Find the last three digits of f(f(f((((f(((((((1 + 1f((1 + f(1)1f(11f((1 + f(1)) + 1) + 11f(f(f(1) + 1)) + 1) + 11) + f(11)) + 1)1f(1 + f(1))))))) + 1) + 1)) + 1) + f(1 + f(1 + f(1)))f(1 + f(1))) + 1) + 1) + 1. -6. Let ABC be a triangle with AC = 5 − √ 3, BC = √ 2. Let A0 , B0 be points outside of triangle ABC such that ∠A0BC = 15◦ , ∠A0CB = 75◦ , ∠ACB0 = 60◦ , ∠B0AC = 30◦ . Given that A0B0 can be expressed in the form a √ b + c, where a, b, c are positive integers and b is not divisible by the square of any prime, find a + b + c. -7. In 4ABC, let AB = 9, BC = 13, and CA = 14. Let G, H, I, and O be the centroid, orthocenter, incenter, and circumcenter of 4ABC, respectively. Let G1 be the centroid of quadrilateral HIOG, and let G0 1 be the isogonal conjugate of G1 with respect to 4ABC. Let line `1 be the polar of G0 1 with respect to the circumcircle of 4ABC, let `2 be the polar of I with respect to 4ABC, and let X be the point of intersection of `1 and `2. Let A0 , B0 , and C 0 be the inverse images of points A, B, and C, with respect to an inversion about X with radius 12. Given that the area of 4ABC can be expressed in the form a √ b c , where b is a positive integer not divisible by the square of any prime and a and c are relatively prime positive integers, find a + b + c. -8. Let n be the largest number for which 1337242010 −8 242010 3n is a positive integer. Find the last three digits of the value of this integer. -9. Given that the root to x − 3 = 0 is m n , where m and n are complex numbers such that m = a + bi, n = c + di, where i is defined as the square root of −4, and a, b, c, d are complex numbers such that a = e+f i, b = g+hi, c = i+ji, d = k+li, where e, f, g, h, j, k, l are positive integers such that g, j, k and l are not divisible by any prime and e is no greater than h, and that (|a 2 + b 2 + c 2 + d 2 + e 2 + f 2 + g 2 + h 2 + i 2 + j 2 + k 2 + l 2 + m2 + n 2 |) 2 = y, where y is a positive integer, find the leftmost three digits of y. -10. Let x be a positive real number satisfying x 3 −4x 2 + 3x−5 = 0. If x = a+ 3 q b+c √ d e + 3 qf+g √ h j k , where a, b, c, d, e, f, g, h, j, k are integers with d, h > 0, gcd(a, k) = gcd(b, c, e) = gcd(f, g, h) = 1, gcd(b, c) and gcd(f, g) not divisible by the cube of any prime. d, h not divisible by the square of any prime, and g > c, find the remainder when abce+d+f +g +h+j +k is divided by 1000. -11. Given that the value of X∞ n=1 n 4 + 40n 3 + 225n 2 + 248n + 4 n6 + 19n5 + 101n4 + 103n3 + 19n2 − 5n + 7 can be expressed in the form p q , where p and q are relatively prime positive integers, find the last three digits of p + q. -12. Find the sum of all positive integers k such that there exist positive integers a, b, c, d, e, f, g, h, i, j, l, m, n, o, p, q, r, s, t, u, v, w, x, y, and z such that Negative AIME 2 wz + h + j − q = 0 (7gk + 7k − 1)(h + j) + h − z = 0 112k(7k − 1)3 (n + 1)2 + 1 − f 2 = 0 2n + p + q + z − e = 0 e 3 (e + 2)(a + 1)2 + 1 − o 2 = 0 (a 2 − 1)y 2 + 1 − x 2 = 0 16r 2 y 4 (a 2 − 1) + 1 − u 2 = 0 n + l + v − y = 0 (a 2 − 1)l 2 + 1 − m2 = 0 ai + 7k − 1 − l − i = 0 �a + u 2 (u 2 − a) 2

− 1

(n + 4dy) 2 + 1 − (x + cu) 2 = 0 p + l(a − n − 1) + b(2an + 2a − n 2 − 2n − 2) − m = 0 q + y(a − p − 1) + s(2ap + 2a − p 2 − 2p − 2) − x = 0 z + pl(a − p) + t(2ap − p 2 − 1) − pm = 0. -13. Solve the following 14 problems. The answer to each will be an integer. Use the following key to convert your numbers into a message which will give you further instructions. If any of your answers have more than one digit, simply use the last digit for converting purposes. 0 1 2 3 4 5 6 7 8 9 D I V A S B O N E R Problem the first: Alex and Anderson independently each pick one of the seven deadly sins to commit. In how many distinct ways can they do this? Problem the second: Let {an} be a sequence such that a1 = 1, an+1 = an + 1. Find a18. Problem the third: We call a scalene triangle with integer side lengths almost isosceles if its two shortest side lengths differ by 1. The smallest (in terms of area) almost isosceles right triangle has side lengths 3,4,5. Find the smallest side length of the next smallest almost isosceles right triangle. Problem the fourth: The Ulam sequence is defined as follows: U1 = 1, U2 = 2, and for n > 2, Un is defined to be the smallest integer that can be expressed as the sum of two distinct earlier terms in exactly one way. Find the eleventh term in this sequence. Problem the fifth: There are four distinguishable hamsters. For lunch, each has the choice of 7 distinguishable meals. How many ways can they eat lunch together? Problem the sixth: Timmy is excited because he is only 1 year away from becoming 18, the legal age of adulthood. How old is Timmy? Problem the seventh: In 1989, L. Ming showed that there are only 4 distinct numbers which are both Fibonacci numbers and triangular numbers. Find the largest of these 4 numbers. Problem the eighth: What is the smallest positive integer n such that 1/2 + 1/3 + 1/7 + 1/n < 1? Problem the ninth: 5 letters are written to different addresses, and 5 matching envelopes are prepared. In how many ways can the letters be placed in the envelopes so that no letter is in a correct envelope? Problem the tenth: Find the largest integer x such that 16x divides (10!)2004 . Negative AIME 3 Problem the eleventh: Let x be the integer two below one hundred, and let y be the number of two-digit primes. Find x − y. Problem the twelfth: Hammy the hamster is 8 years old. If he dies in 3 years, how old (in years) will he be when he dies? Problem the thirteenth: What is the only prime which satisfies the property that it is equal to the sum of the digits (in base ten) of its cube? Problem the fourteenth: How many three-digit numbers (with no leading zeroes) with three distinct digits satisfy the property that if you write the number in reverse, this reversed number is larger? -14. In Billbobland, currency is in the form of bobs, and comes in bill (paper) form. 16 friends named Bobb, Bobby-Bob, Bill, Bobby, Bill-Bob, Bob, Bobby-Bill, Bob-Billy, Billy-Bobby, Billy, Bob-Bill, Billy-Bob, Bob-Bob, Bill-Bobby, Bill-Bobb, and Frank each have some non negative integer amount of bills worth 1 bob each. If the number of bills that Bob-Bill has is equal to the 3 times the number of bills that Billy-Bobby has plus 7 times the number of bobs that Frank has; Bobb’s bob amount is equal to Bobby-Bob’s number of bills plus 2 times the amount of bobs Frank has; Bobby gaining 1337 bills would cause him to have 42 times the amount of bobs as Bill-Bobby; Bob has an amount of bobs equivalent to 67 bobs plus the sum of Bill-Bob’s, Bobby-Bill’s, Bobby’s, and Bob-Billy’s bobs; the combined amount of 350 times Bill-Bobb’s bills plus 10 times Bob-Bob’s bobs plus Bill-Bob’s number of bills is 9001 over Bob-Billy’s number of bills; Bill-Bob has an amount of bills equal to 4 bills plus the amount of bobs that Bob-Billy and Bobby have combined; Bob-Bob’s number of bills is equal to the sum of Billy-Bob and Billy’s number of bobs; the amount of bills that Bobb has plus twice the amount of bobs that Bobby-Bob has is equal to 3 times the amount of bills that Bill has; 3 times the amount of bobs Bobby-Bob has is equal to the sum of Bill’s amount of bills and Bobb’s amount of bills; Frank has no bills nor bobs (which are, admittedly, the same) at all; if Bobby-Bill’s amount of bills had 13 times Frank’s amount of bobs subtracted from it, it would be equal to 8 times the amount of bills that Bob-Billy has plus 2 additional bobs; the sum of Billy-Bobby’s amount of bobs and Billy’s amount of bobs is equal to the sum of Bob-Bill’s amount of bobs and Frank’s amount of bobs; 3 times the amount of bills that Bobby-Bill has plus the amount of bobs that Bill-Bob has is equal to the sum of Bob’s bobs and Bob-Billy’s bobs; 13 times Billy-Bobby’s number of bobs plus 3 times Billy’s number of bills plus 7 times Bob-Bill’s number of bobs equals 120; 2 times Bobby’s number of bills plus 2 times Bill-Bob’s number of bobs plus 17 times Frank’s number of bills equals 4 bobs plus the number of bills Bobby-Bill has; and the 4 times the number of bills Bob-Billy possesses plus the sum of Bobb’s amount of bills, Billy-Bob’s amount of bills, and Bob’s amount of bills equals 3 times Bob-Bob’s number of bobs plus Bill-Bobby’s number of bills plus Bill-Bobb’s number of bobs plus 2 times Bill-Bob’s number of bills plus 3 times Bobby’s number of bobs plus Billy’s number of bills plus 1 lone bob; then, how many bobs do Bobb, Bobby-Bob, Bill, Bobby, Bill-Bob, Bob, Bobby-Bill, Billy-Bobby, Billy, Bob-Bill, Billy-Bob, Bob-Bob, Bill-Bobby, Bill-Bobb, and Frank have combined? Negative AIME 4 -15. Let a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z, be functions such that a(x) = 1x + 2x + 3x + 4x + 5x − 15x + 89 b(x) = 80x 2 + 160x + 80 c(x) = 69x − 69 d(x) = 1 + 3x + 3x 2 + 7 e(x) = 2x 2 − 278x + 9669 f(x) = x − 2 g(x) = 2 − x h(x) = 37x + 13 i(x) = 1 + 33x + 7x 2 + 1337x 3 j(x) = 1337x 1337 k(x) = x l(x) = 1337x m(x) = 1 + 2x 2 + 3x 3 + 4x 4 + 5x 5 n(x) = x + 42 o(x) = x − 42 p(x) = x − 1 j x 1 k q(x) = x − 2 j x 2 k r(x) = � x 5 � s(x) = 42x t(x) = x + 537 u(x) = 13x + 37 v(x) = 964 + 931x w(x) = x 2 x(¯x) = ln e x¯ y(x) = −|x| z(x) = |x| (x) = 70 − x Let f1, f2, f3, f4, f5, f6 be functions such that f1(X) = n(e(v(e(r( (g(o(n(n(a( (g(i(v(e( (y(o(u( (u(p(X))))))))))))))))))))))), f2(X) = n(e(v(e(r( (g(o(n(n(a( (l(e(t( (y(o(u( (d(o(w(n(X)))))))))))))))))))))))), f3(X) = n(e(v(e(r( (g(o(n(n(a( (r(u(n( (a(r(o(u(n(d( (a(n(d( (d(e(s(e(r(t( (y(o(u(X))))))))))))))))))))))))))))))))))))), f4(X) = n(e(v(e(r( (g(o(n(n(a( (m(a(k(e( (y(o(u( (c(r(y(X)))))))))))))))))))))))), f5(X) = n(e(v(e(r( (g(o(n(n(a( (s(a(y( (g(o(o(d(b(y(e(X))))))))))))))))))))))), f6(X) = n(e(v(e(r( (g(o(n(n(a( (t(e(l(l( (a( (l(i(e( (a(n(d( (h(u(r(t( (y(o(u(X))))))))))))))))))))))))))))))))))), and let chorus(x) = f6(f5(f4(f3(f2(f1(x)))))). Negative AIME 5 Let a1 = w(e(r(e( (n(o( (s(t(r(a(n(g(e(r(s( (t(o( (l(o(v(e(42))))))))))))))))))))))))), a2 = y(o(u( (k(n(o(w( (t(h(e( (r(u(l(e(s( (a(n(d( (s(o( (d(o( (i(a1)))))))))))))))))))))))))))))), a3 = a( (f(u(l(l( (c(o(m(m(i(t(m(e(n(t(s( (w(h(a(t( (i(m( (t(h(i(n(k(i(n(g( (o(f(a2)))))))))))))))))))))))))))))))))))))), a4 = y(o(u( (w(o(u(l(d(n(t( (g(e(t( (t(h(i(s( (f(r(o(m( (a(n(y( (o(t(h(e(r( (g(u(y(a3))))))))))))))))))))))))))))))))))))))), a5 = i( (j(u(s(t( (w(a(n(n(a( (t(e(l(l( (y(o(u( (h(o(w( (i(m( (f(e(e(l(i(n(g(a4)))))))))))))))))))))))))))))))))))), a6 = g(o(t(t(a( (m(a(k(e( (y(o(u( (u(n(d(e(r(s(t(a(n(d(a5))))))))))))))))))))))))), a7 = chorus(a6), a8 = w(e(v(e( (k(n(o(w(n( (e(a(c(h( (o(t(h(e(r( (f(o(r( (s(o(( (l(o(n(g(a7)))))))))))))))))))))))))))))))))), a9 = y(o(u(r( (h(e(a(r(t(s( (b(e(e(n( (a(c(h(i(n(g( (b(u(t(a8))))))))))))))))))))))))))), a10 = y(o(u(r(e( (t(o(o( (s(h(y( (t(o( (s(a(y( (i(t(a9))))))))))))))))))))))), a11 = i(n(s(i(d(e( (w(e( (b(o(t(h( (k(n(o(w( (w(h(a(t(s( (b(e(e(n( (g(o(i(n(g( (o(n(a10)))))))))))))))))))))))))))))))))))))), a12 = w(e( (k(n(o(w( (t(h(e( (g(a(m(e( (a(n(d( (w(e(r(e( (g(o(n(n(a( (p(l(a(y( (i(t(a11))))))))))))))))))))))))))))))))))))))), a13 = a(n(d( (i(f( (y(o(u( (a(s(k( (m(e( (h(o(w( (i(m( (f(e(e(l(i(n(g(a12)))))))))))))))))))))))))))))))), a14 = d(o(n(t( (t(e(l(l( (m(e( (y(o(u(r(e( (t(o(o( (b(l(i(n(d( (t(o( (s(e(e(a13))))))))))))))))))))))))))))))))))), a15 = chorus(a14), a16 = chorus(a15), a17 = o(o(h( (g(i(v(e( (y(o(u( (u(p(a16))))))))))))))), a18 = o(o(h( (g(i(v(e( (y(o(u( (u(p(a17))))))))))))))), a19 = n(e(v(e(r( (g(o(n(n(a( (g(i(v(e( (n(e(v(e(r( (g(o(n(n(a( (g(i(v(e(a18))))))))))))))))))))))))))))))))), a20 = g(i(v(e( (y(o(u( (u(p(a19))))))))))), a21 = n(e(v(e(r( (g(o(n(n(a( (g(i(v(e( (n(e(v(e(r( (g(o(n(n(a( (g(i(v(e(a20))))))))))))))))))))))))))))))))), a22 = g(i(v(e( (y(o(u( (u(p(a21))))))))))), a23 = w(e(v(e( (k(n(o(w(n( (e(a(c(h( (o(t(h(e(r( (f(o(r( (s(o(( (l(o(n(g(a22)))))))))))))))))))))))))))))))))), a24 = y(o(u(r( (h(e(a(r(t(s( (b(e(e(n( (a(c(h(i(n(g( (b(u(t(a23))))))))))))))))))))))))))), a25 = y(o(u(r(e( (t(o(o( (s(h(y( (t(o( (s(a(y( (i(t(a24))))))))))))))))))))))), a26 = i(n(s(i(d(e( (w(e( (b(o(t(h( (k(n(o(w( (w(h(a(t(s( (b(e(e(n( (g(o(i(n(g( (o(n(a25)))))))))))))))))))))))))))))))))))))), a27 = w(e( (k(n(o(w( (t(h(e( (g(a(m(e( (a(n(d( (w(e(r(e( (g(o(n(n(a( (p(l(a(y( (i(t(a26))))))))))))))))))))))))))))))))))))))), a28 = i( (j(u(s(t( (w(a(n(n(a( (t(e(l(l( (y(o(u( (h(o(w( (i(m( (f(e(e(l(i(n(g(a27)))))))))))))))))))))))))))))))))))), a28 = g(o(t(t(a( (m(a(k(e( (y(o(u( (u(n(d(e(r(s(t(a(n(d(a28))))))))))))))))))))))))), a30 = chorus(a29), a31 = chorus(a30), a32 = chorus(a31) Compute t(h(e( (q(u(i(c(k( (b(r(o(w(n( (f(o(x( (j(u(m(p(e(d( (o(v(e(r( (t(h(e( (l(a(z(y( (d(o(g(a32))))))))))))))))))))) ))))))))))))))))))))))). Happy April Fools’ Day! As you may have noticed, the problems aren’t quite... standard. But each one is completely possible. Some of these problems are really wacky, and we hope you’ll take some time to twist your mind, actually do the test, and participate in a strange competition. We’ll recognize the high scorers as well as any notable responses we receive. Have fun!