2012 AMC 10B Problems/Problem 8

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Problem 8

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

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

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

$(x-2)^2=16$

$x=6,-2$

What is the sum of all integer solutions

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

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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