2021 April MIMC 10 Problems/Problem 7

Find the least integer $k$ such that $838_k=238_k+1536$ where $a_k$ denotes $a$ in base-$k$.

$\textbf{(A)} ~12 \qquad\textbf{(B)} ~13 \qquad\textbf{(C)} ~14 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16$

Solution

We can express $838_k=238_k+1536$ to be $8k^2+3k+8=2k^2+3k+8+1536$ in base $10$. Notice that $3k+8$ cancels out as it is on both sides of the equation, and we can move $2k^2$ to the left side of the equation. This results in $6k^2=1536$, $k^2=256$. Since $k$ is a positive integer, $k=\fbox{\textbf{(E)} 16}$.

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