# 2012 AMC 12B Problems/Problem 11

## Problem

In the equation below, $A$ and $B$ are consecutive positive integers, and $A$, $B$, and $A+B$ represent number bases: $$132_A+43_B=69_{A+B}.$$ What is $A+B$?

$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 17$

## Solution 1

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 $A+B = \boxed{\textbf{(C)}\ 13}$

We can eliminate answer choice $\textbf{(A)}$ because you can't have a $9$ in base $9$. Now we know that A and B are consecutive, so we can just test answers. You will only have to test at most $8$ cases. Eventually, after testing a few cases, you will find that $A=6$ and $B=7$. The solution is $\boxed{\mathbf{(C)}\ 13}$.