Difference between revisions of "2012 AMC 10B Problems/Problem 8"
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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> | <math> \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25 </math> | ||
| − | == | + | == Solution == |
<math>(x-2)^2</math> = perfect square. | <math>(x-2)^2</math> = perfect square. | ||
| − | 1< perfect square< 25 | + | 1 < perfect square < 25 |
Perfect square can equal: 4, 9, or 16 | Perfect square can equal: 4, 9, or 16 | ||
| − | Solve for x: | + | Solve for <math>x</math>: |
<math>(x-2)^2=4</math> | <math>(x-2)^2=4</math> | ||
| Line 31: | Line 31: | ||
<math>x=6,-2</math> | <math>x=6,-2</math> | ||
| − | + | The sum of all integer solutions is | |
<math>4+5+6+0+(-1)+(-2)=\boxed{\textbf{(B)} 12}</math> | <math>4+5+6+0+(-1)+(-2)=\boxed{\textbf{(B)} 12}</math> | ||
Latest revision as of 17:08, 13 July 2021
Problem 8
What is the sum of all integer solutions to 
?
Solution
= perfect square.
1 < perfect square < 25
Perfect square can equal: 4, 9, or 16
Solve for 
:
and
and
The sum of all integer solutions is
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
