PUMaC 2010 · 数论（B 组） · 第 8 题

PUMaC 2010 — Number Theory (Division B) — Problem 8

Let N be the number of (positive) divisors of 2010 ending in the digit 2. What is the remainder when N is divided by 2010? 1

解析

Let N be the number of (positive) divisors of 2010 ending in the digit 2. What is the remainder when N is divided by 2010? a b c Solution: All such divisors are of the form 2 3 67 with 1 ≤ a ≤ 2010 and 0 ≤ b, c, ≤ 2010. Moreover, we have a − b + c ≡ 1( mod 4). 503, 503, 502, and 502 are the numbers of ways a can be congruent to 1 , 2 , 3 , or 0( mod 4), respectively. b and c can each be congruent to 1 , 2 , 3 , and 0 mod 4 in 503, 503, 502, and 503 ways. 2 2 2 When a ≡ 1, we have b − c ≡ 0 which can happen 3 · 503 + 502 = (502 + 503) + 2(503) ≡ 2 1005 + 1006 ≡ 1005 + 1006 ≡ 1 mod 4 . This gives us 503+2010k possibilities for some integer 2 k. For each of the other possibilities, namely, a ≡ 2 , 3 , or 0 mod 4, there are 2 · 503 + 2 · 503 · 502 = 2 · (503 + 502) · 503 = 2010 · 503 ways. This is zero mod 2010, so the answer is just 503. 2