Question

所有满足的整数解之和是多少？ $1 < (x-2)^2 < 25?$

What is the sum of all integer solutions to $1 < (x-2)^2 < 25?$

$10$

$12$

$15$

$19$

$25$

答案：B

知识点：不等式绝对值对称性

难度评级：1140

解答：

若 $x=2+k$ 是一个解，则 $x=2-k$ 也是解，因为这两个解的和为 $4$ 。因此，所有整数解之和等于四乘以可行的正数 $k$ 的个数。 $((2+k)-2)^2 = ((2-k)-2)^2$

正数 $k$ 必须满足共有 $3$ 个，即 $k=2,3,4$ 。 $1 < k^2 < 25,$

所以总和为 $4\cdot3=12$ 。

所以正确答案是 B。

Suppose we have $x=2+k$ as a solution. Then, $x=2-k$ would also be a solution as $((2+k)-2)^2 = ((2-k)-2)^2$ The sum of these two solutions would be $4.$ Thus, the sum of all integer solutions to the above equation is four times the number of positive $k$ 's that work.

To find the number of $k$ 's, we need to find the number of positive solutions to: $1 < k^2 < 25,$ which would be $3,$ as $k=2,3,4.$

Therefore, the sum of all the solutions is $4\cdot3=12.$

Thus, the correct answer is B .

2012 AMC 10B 第 8 题

答案：B

难度评级：1140

解答：

其他年份的第 8 题