Difference between revisions of "2012 AMC 10B Problems/Problem 8"

Revision as of 09:52, 28 August 2020

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

$(x-2)^2=16$

$x=6,-2$

What is the sum of all integer solutions

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

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 3: / Line 3: @@
 What is the sum of all integer solutions to <math>1<(x-2)^2<25</math>?
-<math> \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\25 </math>
+<math> \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25 </math>
-[[2012 AMC 10B Problems/Problem 8|Solution]]
+== Solution ==
-== Solutions ==
 <math>(x-2)^2</math> = perfect square.
@@ Line 22: / Line 17: @@
 <math>(x-2)^2=4</math>
-<math>x=4</math>
+<math>x=4,0</math>
 and
@@ Line 28: / Line 23: @@
 <math>(x-2)^2=9</math>
-<math>x=5</math>
+<math>x=5,-1</math>
 and
@@ Line 34: / Line 29: @@
 <math>(x-2)^2=16</math>
-<math>x=6</math>
+<math>x=6,-2</math>
 ''What is the sum of all integer solutions''
-<math>4+5+6=\boxed{15}</math>
+<math>4+5+6+0+(-1)+(-2)=\boxed{\textbf{(B)} 12}</math>
-OR
+==See Also==
-<math> \textbf{(C)}</math>
+{{AMC10 box|year=2012|ab=B|num-b=7|num-a=9}}
+{{MAA Notice}}