HMMT 十一月 2021 · 冲刺赛 · 第 28 题

28. [15] Find the smallest positive integer n such that the divisors of n can be partitioned into three sets with equal sums. Proposed by: Maxim Li Answer: 120 3 Solution: I claim the answer is 120. First, note that 120 = 2 · 3 · 5, so the sum of divisors is (1 + 2 + 4 + 8)(1 + 3)(1 + 5) = 15 · 4 · 6 = 360. Thus, we need to split the divisors into groups summing to 120. But then we can just take { 120 } , { 20 , 40 , 60 } , { 1 , 2 , 3 , 4 , 5 , 6 , 8 , 10 , 12 , 15 , 24 , 30 } . Thus, 120 works. Now we need to show 120 is the lowest. Let s ( n ) be the sum of divisors. Since n will be in one of the piles, we need s ( n ) ≥ 3 n . First, we claim that n must have at least 3 distinct prime divisors. Surely, a b if it had 2 distinct prime divisors, say p and q, so that n = p q , then the sum of divisors is ( ) ( ) 1 1 1 1 2 a 2 b a b (1 + p + p + ... + p )(1 + q + q + ... + q ) = p q 1 + + ... + 1 + + ... + . a b p p q q 1 1 However, the expression 1 + + ... + is maximized when p is minimized, and further, as a is finite a p p p 1 must be at most = . Thus, the sum of divisors is less than 1 p − 1 1 − p p q 3 a b p q ≤ n · 2 · = 3 n. p − 1 q − 1 2 Thus, n can’t have 2 distinct prime divisors and must have at least 3 distinct prime divisors. As we already discovered 120 works, we need not worry about 4 distinct prime divisors, as the value of n would be at least 2 · 3 · 5 · 7 = 210 . We now work through the numbers with 3 distinct divisors. If 2 is not one of them, then the only number that works is 105 = 3 · 5 · 7 , which has a sum of di- visors that is not large enough. Therefore, 2 must be a prime divisor of n. Additionally, if 3 is not a divisor, then our options are 2 · 5 · 7 and 2 · 5 · 11 , which also do not work. Therefore, 3 must also be a prime divisor. Then, if 5 is not a prime divisor, then if n is 2 · 3 · p, it has a sum of divisors of p +1 3 4 (1 + 2)(1 + 3)(1 + p ) = n · · · , which is only at least 3 n if p is exactly 2 , which is not feasible. 2 3 p p 2 7 4 Additionally, if we use 2 , then the sum of divisors is (1 + 2 + 4)(1 + 3)(1 + p ) = n · · · , so 4 3 p +1 p +1 9 2 2

= ⇒ p < 4 . 5 , which also can’t happen. Further, we can’t have 3 be a divisor of n as 2 · 3 · 5 p 7 is the only value less than 120 with this, and that also does not work. Lastly, we just need to check p +1 p 15 4 5 3 2 · 3 · p, which has a sum of divisors of (1+2+4+8)(1+3)(1+ p ) = n · · · = n · · , so p = 5 and 8 3 p 2 p +1 that works. This means that n = 120 is the smallest value for which s ( n ) ≥ 3 n, and thus is our answer.