2012 AMC 10B Problem 8

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Concepts:inequalityabsolute valuesymmetry

Difficulty rating: 1140

8.

What is the sum of all integer solutions to 1<(x2)2<25?1 < (x-2)^2 < 25?

10 10

12 12

15 15

19 19

25 25

Solution:

Suppose we have x=2+kx=2+k as a solution. Then, x=2kx=2-k would also be a solution as ((2+k)2)2=((2k)2)2((2+k)-2)^2 = ((2-k)-2)^2 The sum of these two solutions would be 4.4. Thus, the sum of all integer solutions to the above equation is four times the number of positive kk's that work.

To find the number of kk's, we need to find the number of positive solutions to: 1<k2<25,1 < k^2 < 25, which would be 3,3, as k=2,3,4.k=2,3,4.

Therefore, there are a total of 43=124\cdot 3=12 solutions.

Thus, the correct answer is B .

Problem 8 in Other Years