2012 AMC 10B 第 8 题

Question

求满足下列不等式的所有整数解之和：

1 < (x-2)^2 < 25

What is the sum of all integer solutions to the following inequality?

1 < (x-2)^2 < 25

10

12

15

19

25

把条件看成未知数到

2

的距离。

Solve by distance from

2

整数解满足

|x-2|=2,3,4

。

Integer solutions have

|x-2|=2,3,4

若

x=2+k

是一个解，则

x=2-k

也是解，因为

((2+k)-2)^2 = ((2-k)-2)^2

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。

正数

k

必须满足

1 < k^2 < 25\text{，}

3

个，即

k=2,3,4

。

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 .

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