2012 AMC 12B Problems/Problem 11

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.

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2012 AMC 12B Problems/Problem 11

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