Difference between revisions of "2012 AMC 12B Problems/Problem 11"

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Either <math>B = A + 1</math> or <math>B = A - 1</math>, so either <math>A^2 - 5A - 6, B = A + 1</math> or <math>A^2 - 5A - 2, B = A - 1</math>. The second case has no integer roots, and the first can be re-expressed as <math>(A-6)(A+1) = 0, B = A + 1</math>. Since A must be positive, <math>A = 6, B = 7</math> and <math>A+B = 13</math>; C.
 
Either <math>B = A + 1</math> or <math>B = A - 1</math>, so either <math>A^2 - 5A - 6, B = A + 1</math> or <math>A^2 - 5A - 2, B = A - 1</math>. The second case has no integer roots, and the first can be re-expressed as <math>(A-6)(A+1) = 0, B = A + 1</math>. Since A must be positive, <math>A = 6, B = 7</math> and <math>A+B = 13</math>; C.
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== See Also ==
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{{AMC12 box|year=2012|ab=B|num-b=10|num-a=12}}

Revision as of 22:20, 12 January 2013

Problem

In the equation, 132 (base A) + 43 (base B) = 69 (base A+B), A and B are consecutive integers. What is A+B?


Solution

Change the equation to base 10: \[A^2 + 3A +2 + 4B +3= 6A + 6B + 9\] \[A^2 - 3A - 2B - 4=0\]

Either $B = A + 1$ or $B = A - 1$, so either $A^2 - 5A - 6, B = A + 1$ or $A^2 - 5A - 2, B = A - 1$. The second case has no integer roots, and the first can be re-expressed as $(A-6)(A+1) = 0, B = A + 1$. Since A must be positive, $A = 6, B = 7$ and $A+B = 13$; C.

See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AMC 12 Problems and Solutions
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