PUMaC 2021 · 团队赛 · 第 3 题

3 . This means that M is only divisible by one of 5 , 7 , and that this power needs to be either 1 or 2 . 672 671 2 With a sum of primes being 2023 , we see that our candidates are M = 3 · 7 and M = 3 · 5 . 2 671 2 But 3 · 7 < 5 , meaning that M = 3 · 5 . We now need to find m. Notice that m must be divisible by at least 3 primes, since m is odd. 2 We claim that m = 3 · 2017 . We see that this lies in S and is odd. Suppose that we have an integer n whose largest prime is p. Then, observe that n ≥ p (2023 − p ) , since a product of integers that are at least 2 is larger than their sum. However, notice that the largest primes less than 2023 that is at most 9 away from 2014 is 2017; any other choice of largest prime yields p (2023 − p ) is larger than 20000 , whereas 9 · 2017 is smaller than this. 2 669 2 Therefore, m = 3 · 2017 , and so M/m = 3 5 / 2017 , meaning that our answer is equal to 3 + 669 + 5 + 2 + 2017 − 1 = 2695 .