2012 AMC 10B Problems/Problem 8

Problem 8

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

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25$

$(x-2)^2$ = perfect square.

1 < perfect square < 25

Perfect square can equal: 4, 9, or 16

Solve for $x$ :

$(x-2)^2=4$

$x=4,0$

and

$(x-2)^2=9$

$x=5,-1$

and

$(x-2)^2=16$

$x=6,-2$

The sum of all integer solutions is

$4+5+6+0+(-1)+(-2)=\boxed{\textbf{(B)} 12}$

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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