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

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What is the sum of all integer solutions to <math>1<(x-2)^2<25</math>?
 
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>
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<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]]
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== Solution ==
 
 
 
 
 
 
 
 
== Solutions ==
 
  
 
<math>(x-2)^2</math> = perfect square.
 
<math>(x-2)^2</math> = perfect square.
  
1< perfect square< 25
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1 < perfect square < 25
  
 
Perfect square can equal: 4, 9, or 16
 
Perfect square can equal: 4, 9, or 16
  
Solve for x:
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Solve for <math>x</math>:
  
 
<math>(x-2)^2=4</math>  
 
<math>(x-2)^2=4</math>  
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<math>x=6,-2</math>
 
<math>x=6,-2</math>
  
''What is the sum of all integer solutions''
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The sum of all integer solutions is
  
<math>4+5+6+0+(-1)+(-2)=\boxed{12}</math>
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<math>4+5+6+0+(-1)+(-2)=\boxed{\textbf{(B)} 12}</math>
  
OR
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==See Also==
  
<math> \textbf{(B)}</math>
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{{AMC10 box|year=2012|ab=B|num-b=7|num-a=9}}
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{{MAA Notice}}

Latest revision as of 17:08, 13 July 2021

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
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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