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

Since they are 2 consecutive integers, we can say B=a+1. Now changing to base 10: a^2 + 3a +2 + 4a+4 +3= 12a + 6 +9. Now combining like terms, A^2 + 7a +9 = 12a + 15. so, a^2 - 5a -6= 0. By factoring, we get, (a-6)x(a+1)=0 and since no base is negative, a=6 and b=7, so a+b= 13; C.