Problem:
In the equation below, A and B are consecutive positive integers, and A,B, and A+B represent number bases:
132A+43B=69A+B
What is A+B ?
Answer Choices:
A. 9
B. 11
C. 13
D. 15
E. 17
Solution:
First assume B=A−1. By the definition of number bases,
A2+3A+2+4(A−1)+3=6(A+A−1)+9
Simplifying yields A2−5A−2=0, which has no integer solutions.
Next assume B=A+1. In this case
A2+3A+2+4(A+1)+3=6(A+A+1)+9
which simplifies to A2−5A−6=(A−6)(A+1)=0. The only positive solution is A=6. Letting A=6 and B=7 in the original equation produces 1326+437=6913, or 56+31=87, which is true. The required sum is A+B =13.
The problems on this page are the property of the MAA's American Mathematics Competitions