2012 AMC 12B Problem 11

Attempt Problem 11 of the 2012 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 12B solutions, or check the answer key.

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11.

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?

9

11

13

15

17

Rewrite each numeral in terms of its base, e.g.

132_A = A^2+3A+2

Try both

B=A-1

and

B=A+1

and solve the resulting quadratic in

A

Writing the numerals out,

132_A=A^2+3A+2,

43_B=4B+3,

and

69_{A+B}=6(A+B)+9.

With

B=A+1,

the equation becomes

A^2+3A+2

+4(A+1)+3

=6(2A+1)+9,

which simplifies to

(A-6)(A+1)=0.

The positive solution is

A=6,

B=7.

(The case

B=A-1

gives

A^2-5A-2=0,

which has no integer solution.)

Therefore

A+B=13.

Thus, the correct answer is C.

Problem 11 in other years