1971 AHSME Problems/Problem 11

Problem

The numeral $47$ in base a represents the same number as $74$ in base $b$ . Assuming that both bases are positive integers, the least possible value of $a+b$ written as a Roman numeral, is

$\textbf{(A) }\mathrm{XIII}\qquad \textbf{(B) }\mathrm{XV}\qquad \textbf{(C) }\mathrm{XXI}\qquad \textbf{(D) }\mathrm{XXIV}\qquad \textbf{(E) }\mathrm{XVI}$

Solution

$4a+7=7b+4\implies 7b=4a+3\implies b=\frac{4a+3}{7}$ . $4a+3\equiv 0\implies 4a\equiv 4\implies a\equiv1 \pmod{7}$ . The smallest possible value of $a$ is $8$ . Then, $b=\frac{4\cdot8+3}{7}=\frac{35}{7}=5$ . However, the digit $7$ is not valid in base $5$ , so we have to try a larger value. $a=8+7=15$ , gives a value of $\frac{15*4+3}{7}=9$ , for $b$ , which is valid.

$15+9=24$ , which is $\mathrm{XXIV}$ as a roman numeral, and thus the answer is $\boxed{\textbf{(D) }\mathrm{XXIV}}.$

1971 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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1971 AHSME Problems/Problem 11

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