HMMT 二月 2006 · 冲刺赛 · 第 2 题
HMMT February 2006 — Guts Round — Problem 2
题目详情
英文原题
- [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
英文解析
- Multiplying the three yields ( XY Z ) = 64 from which XY Z = ± 8. Since we aregiven 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
5 5 52
compute a + b + c .
n n n
Defining S = a + b + c , we have S = 4 S + S − S for n ≥ 0. Evidentlyn n +3 n +2 n +1 n
S = 3 , S = 4 , S = ( a + b + c ) − 2( ab + bc + ca ) = 18. Then S = 4 · 18 + 4 − 3 = 73,2
0 1 2 3
5 51
S = 4 · 73 + 18 − 4 = 306, and S = 4 · 306 + 73 − 18 = 1279. Since | b + c | < , we
4 5
5 512
conclude that | S − a | < , or that 1279 is the integer nearest to a .
25