2012 AMC 10B Problems/Problem 8

Problem 8

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

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

Solution

$(x-2)^2$ = 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

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

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AMC 10 Problems and Solutions

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