1967 AHSME Problems/Problem 16

Problem

Let the product $(12)(15)(16)$ , each factor written in base $b$ , equals $3146$ in base $b$ . Let $s=12+15+16$ , each term expressed in base $b$ . Then $s$ , in base $b$ , is

$\textbf{(A)}\ 43\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 47$

Solution

Converting everything into base $10$ , we have $(b + 2)(b+5)(b+6) = 3b^3 + b^2 + 4b + 6$ . Looking ahead, the constant term of the polynomial will be $2*5*6 -6 = 54$ . By the Rational Root Theorem, the only possible integer roots are $1, 2, 3, 6, 9, 18, 27, 54$ . Bases $1, 2, 3, 6$ do not have a $6$ as a digit. Testing $b=9$ gives a solution that works.

Therefore, we are working in base $9$ . Adding the units place in base $9$ , $2 + 5 + 6 = 14_9$ , so we carry the $1$ to get a total of $44_9$ , which is option $\fbox{B}$ .

1967 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1967 AHSME Problems/Problem 16

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