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

Solution 2 (Answer Choices)

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 $\boxed{\mathbf{(C)}\ 13}$ .

~kempwood

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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2012 AMC 12B Problems/Problem 11

Contents

Problem

Solution 1

Solution 2 (Answer Choices)

See Also