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AIME 1997 · 第 1 题

AIME 1997 — Problem 1

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?

解析

Solution

Notice that all odd numbers can be obtained by using (a+1)2−a2=2a+1,(a+1)^2-a^2=2a+1, where aa is a nonnegative integer. All multiples of 44 can be obtained by using (b+1)2−(b−1)2=4b(b+1)^2-(b-1)^2 = 4b, where bb is a positive integer. Numbers congruent to 2(mod4)2 \pmod 4 cannot be obtained because squares are 0,1(mod4).0, 1 \pmod 4. Thus, the answer is 500+250=750.500+250 = \boxed{750}.