1971 AHSME Problems/Problem 11

Problem 11

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

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


$4a+7=7b+4\implies 7b=4a+3\implies b=\frac{4a+3}{7}$. $4a+3\equiv 0\implies 4a\equiv 4\implies a\equiv1 \pmod{7}$. 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 $(\textbf{D})$