HMMT 十一月 2011 · 冲刺赛 · 第 17 题

HMMT November 2011 — Guts Round — Problem 17

[ 10 ] For a positive integer n , let p ( n ) denote the product of the positive integer factors of n . Determine the number of factors n of 2310 for which p ( n ) is a perfect square.

解析

[ 10 ] For a positive integer n , let p ( n ) denote the product of the positive integer factors of n . Determine the number of factors n of 2310 for which p ( n ) is a perfect square. Answer: 27 Note that 2310 = 2 × 3 × 5 × 7 × 11. In general, we see that if n has d ( n ) positive d n 2 integer factors, then p ( n ) = n since we can pair factors ( d, ) which multiply to n . As a result, p ( n ) d is a square if and only if n is a square or d is a multiple of 4. Thus, because 2310 is not divisible by the square of any prime, we claim that for integers n dividing 2310, p ( n ) is even if and only if n is not prime. Clearly, p ( n ) is simply equal to n when n is prime, and p (1) = 1, so it suffices to check the case when n is composite. Suppose that n = p p · · · p , where 1 2 k k k > 1 and { p , . . . , p } is some subset of { 2 , 3 , 5 , 7 , 11 } . Then, we see that n has 2 factors, and that 1 k k 4 | 2 , so p ( n ) is a square. Guts Round 5 Since 2310 has 2 = 32 factors, five of which are prime, 27 of them have p ( n ) even.