2007 AMC 8 第 19 题

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

设连续整数为

x

和

x+1

Let the consecutive integers be

x

and

x+1

.

它们平方差可化简为这两个整数的和

The difference of their squares simplifies to the sum of the two integers.

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}\text{。}

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.

其他年份的第 19 题