HMMT 二月 2006 · 冲刺赛 · 第 2 题

HMMT February 2006 — Guts Round — Problem 2

[5] Compute the positive integer less than 1000 which has exactly 29 positive proper divisors. (Here we refer to positive integer divisors other than the number itself.)

解析

Multiplying the three yields ( XY Z ) = 64 from which XY Z = ± 8. Since we are given X > 0, multiplying the last equation by X we have 2 X = XY Z = ± 8. Evidently XY Z = 8 from which X = 4 , Y = − 1 , Z = − 2. We conclude that a, b, c are the roots 3 2 3 2 of the polynomial P ( t ) = t − 4 t − t + 1. Thus, P ( a ) = a − 4 a − a + 1 = 0, and 3 11 also P ( b ) = P ( c ) = 0. Now since P (1 / 2) = − , P (0) = 1 and P ( − 2 / 3) = − , we 8 27 5 5 1 conclude that − 2 / 3 < c < 0 < b < 1 / 2 < a . It follows that | b + c | < . Thus, we 2 5 5 5 compute a + b + c . n n n Defining S = a + b + c , we have S = 4 S + S − S for n ≥ 0. Evidently n n +3 n +2 n +1 n 2 S = 3 , S = 4 , S = ( a + b + c ) − 2( ab + bc + ca ) = 18. Then S = 4 · 18 + 4 − 3 = 73, 0 1 2 3 1 5 5 S = 4 · 73 + 18 − 4 = 306, and S = 4 · 306 + 73 − 18 = 1279. Since | b + c | < , we 4 5 2 1 5 5 conclude that | S − a | < , or that 1279 is the integer nearest to a . 5 2