PUMaC 2012 · 数论（B 组） · 第 2 题

PUMaC 2012 — Number Theory (Division B) — Problem 2

[ 3 ] Let M be the smallest positive multiple of 2012 that has 2012 divisors. Suppose M can be written as n ∏ a k p k k =1 where the p ’s are distinct primes and the a ’s are positive integers. Find k k n ∑ ( p + a ) . k k k =1 ( ) 2 12 12 12

解析

[ 3 ] Let M be the smallest positive multiple of 2012 that has 2012 divisors. Suppose M can be written as n ∏ a k p k k =1 where the p ’s are distinct primes and the a ’s are positive integers. Find k k n ∑ ( p + a ) . k k k =1 2 Solution: The prime factorization of 2012 is 2 × 503. We see that the solution must have a factor of 2 greater than or equal to 4 and a factor of 503. One of hte primes must be raised to 502 the 503rd power. Because 2 is the smallest prime number, M must be divisible by 2 in order 502 1 1 to minimize its value. We then see that 2 × 3 × 503 has 503 × 2 × 2 = 2012 factors and must be the minimal number satisfying the conditions. So we have 2+502+3+1+503+1 = 1012 . Problem contributed by Andy Loo. ( ) 2 12 12 12