2012 AMC 10B Problems/Problem 8: Difference between revisions

Latest revision as of 17:54, 19 October 2025

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
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

@@ Line 5: / Line 5: @@
 <math> \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25 </math>
-== Solutions ==
+== Solution ==
 <math>(x-2)^2</math> = perfect square.
-< perfect square< 25
+< perfect square < 25
 Perfect square can equal: 4, 9, or 16
-Solve for x:
+Solve for <math>x</math>:
 <math>(x-2)^2=4</math>
@@ Line 31: / Line 31: @@
 <math>x=6,-2</math>
-''What is the sum of all integer solutions''
+The sum of all integer solutions is
-<math>4+5+6+0+(-1)+(-2)=\boxed{12}</math>
+<math>4+5+6+0+(-1)+(-2)=\boxed{\textbf{(B)} 12}</math>
-OR
-<math> \textbf{(B)}</math>
 ==See Also==
@@ Line 43: / Line 39: @@
 {{AMC10 box|year=2012|ab=B|num-b=7|num-a=9}}
 {{MAA Notice}}
+[[Category: Introductory Algebra Problems]]