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Difference between revisions of "2021 April MIMC 10 Problems/Problem 7"

(Created page with "Find the least integer <math>k</math> such that <math>838_k=238_k+1536</math> where <math>a_k</math> denotes <math>a</math> in base-<math>k</math>. <math>\textbf{(A)} ~12 \qq...")
 
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==Solution==
 
==Solution==
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To be Released on April 26th.
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We can express <math>838_k=238_k+1536</math> to be <math>8k^2+3k+8=2k^2+3k+8+1536</math> in base <math>10</math>. Notice that <math>3k+8</math> cancels out as it is on both sides of the equation, and we can move <math>2k^2</math> to the left side of the equation. This results in <math>6k^2=1536</math>, <math>k^2=256</math>. Since <math>k</math> is a positive integer, <math>k=\fbox{\textbf{(E)} 16}</math>.

Latest revision as of 12:34, 26 April 2021

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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