2012 AMC 10B Problems/Problem 8

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Problem 8

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

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\25$ (Error compiling LaTeX. Unknown error_msg)

2012 AMC 10B Problems/Problem 8

Solutions

$(x-2)^2$ = 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

$(x-2)^2=16$

$x=6,-2$

What is the sum of all integer solutions

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

OR

$\textbf{(B)}$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png