1967 AMC 12 Problem 16

Attempt Problem 16 of the 1967 AMC 12 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 1967 AMC 12 solutions, or check the answer key.

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16.

Let the product

(12)(15)(16),

each factor written in base

b,

equal

3146

in base

b.

Let

s=12+15+16,

each term expressed in base

b.

Then

s,

in base

b,

is:

43

44

45

46

47

Translate the three factors and

3146_b

into polynomials in

b

The resulting cubic has the valid base

b=9

The product equation is

\begin{aligned} &(b+2)(b+5)(b+6)\\ &\qquad=3b^3+b^2+4b+6. \end{aligned}

It simplifies to

\begin{aligned} b^3-6b^2-24b-27&=0,\\ (b-9)(b^2+3b+3)&=0. \end{aligned}

Thus the valid base is

b=9.

The required sum is

3b+13=4b+4,

which is

44_b.

Therefore, the correct answer is B.

Problem 16 in other years