1967 AHSME Problems/Problem 16

Revision as of 01:59, 13 July 2019 by Talkinaway (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $\fbox{B}$.

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS