PUMaC 2013 · 数论（A 组） · 第 8 题

PUMaC 2013 — Number Theory (Division A) — Problem 8

[ 8 ] Find the number of primes p between 100 and 200 for which x + y ≡ 2013 (mod p ) has a solution in integers x and y . 1

解析

[ 8 ] Find the number of primes p between 100 and 200 for which x + y ≡ 2013 (mod p ) has a solution in integers x and y . 11 Solution Note that if p 6 = 1 (mod 11), then x will cycle over all residues modulo p , and thus there will always be a solution. It is thus left to check the primes that are 1 modulo 11. Of the numbers between 100 and 200 that are prime, 199 is the only one that is 1 (mod 11). It remains to conclude that there is a solution to this equation mod 199. For example, one solution is x = 5 , y = 46. Therefore the answer is simply the number of primes between 100 and 200, which is 21. 3