HMMT 二月 2022 · 冲刺赛 · 第 10 题

HMMT February 2022 — Guts Round — Problem 10

[7] A positive integer n is loose it has six positive divisors and satisfies the property that any two positive divisors a < b of n satisfy b ≥ 2 a . Compute the sum of all loose positive integers less than 100.

解析

[7] A positive integer n is loose it has six positive divisors and satisfies the property that any two positive divisors a < b of n satisfy b ≥ 2 a . Compute the sum of all loose positive integers less than 100. Proposed by: Daniel Zhu Answer: 512 Solution: Note that the condition in the problem implies that for any divisor d of n , if d is odd then d all other divisors of n cannot lie in the interval [ ⌈ ⌉ , 2 d − 1]. If d is even, then all other divisors cannot 2 d lie in the interval [ + 1 , 2 d − 1]. 2 5 2 5 We first find that n must be of the form p or p q for primes p and q . If n = p , the only solution is when p = 2 and n = 32. 2 2 Otherwise, n = p q . Since 100 > n > p , so p ≤ 7. Now we can do casework on p . When p = 2, we find that q cannot lie in [2 , 3] or [3 , 7], so we must have q ≥ 11. All such values for q work, giving solutions n = 44 , 52 , 68 , 76 , 92. When p = 3, we find that q cannot lie in [2 , 5] or [5 , 17], so we must have that q ≥ 19, so there are no solutions in this case. When p = 5 or p = 7, the only solution occurs when q = 2 (since otherwise n > 100). This gives us the solutions n = 50 and n = 98. Adding these values of n gives 512.