Question

选择两个连续正整数，它们的和小于 $100$ 。把这两个整数分别平方，然后求平方差。下列哪一个可能是这个差？

Pick two consecutive positive integers whose sum is less than $100.$ Square both of those integers and then find the difference of the squares. Which of the following could be the difference?

$2$

$64$

$79$

$96$

$131$

答案：C

知识点：平方差奇偶性

难度评级：1430

解答：

设连续正整数为 $x$ 和 $x+1$ 。它们的和为 $2x+1$ ，小于 $100$ 。

它们的平方差为 $\begin{gathered} (x+1)^2-x^2 \\ = (x+1+x)(x+1-x) \\ = 2x+1. \end{gathered}$

所以这个差必须是小于 $100$ 的奇数。选项中只有 $79$ 可能，它由 $39$ 和 $40$ 得到。

所以正确答案是 C。

Let the consecutive positive integers be $x$ and $x+1$ . Their sum is $2x+1$ , which is less than $100$ .

The difference of their squares is $\begin{gathered} (x+1)^2-x^2 \\ = (x+1+x)(x+1-x) \\ = 2x+1. \end{gathered}$

So the difference must be an odd number less than $100$ . Among the choices, $79$ is the only possibility, and it occurs for $39$ and $40$ .

Thus, C is the correct answer.

2007 AMC 8 第 19 题

答案：C

难度评级：1430

解答：

其他年份的第 19 题