Difference between revisions of "1965 AHSME Problems/Problem 15"

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== Solution ==
 
== Solution ==
  
We begin by converting both <math>25_b</math> and <math>52_b</math> to base <math>10</math>. <math>25_b = 2b+5</math> in base <math>10</math> and <math>52_b = 5b+2</math> base <math>10</math>. The problem tells us that <math>5b+2 = 4b+10</math>, yielding <math>\boxed{8}</math> as our final answer.
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We begin by converting both <math>25_b</math> and <math>52_b</math> to base <math>10</math>. <math>25_b = 2b+5</math> in base <math>10</math> and <math>52_b = 5b+2</math> base <math>10</math>. The problem tells us that <math>5b+2 = 4b+10</math>, yielding <math>\boxed{\textbf{(B) }8}</math> as our final answer.

Revision as of 12:11, 18 July 2024

Problem

The symbol $25_b$ represents a two-digit number in the base $b$. If the number $52_b$ is double the number $25_b$, then $b$ is:

$\textbf{(A)}\ 7 \qquad  \textbf{(B) }\ 8 \qquad  \textbf{(C) }\ 9 \qquad  \textbf{(D) }\ 11 \qquad  \textbf{(E) }\ 12$

Solution

We begin by converting both $25_b$ and $52_b$ to base $10$. $25_b = 2b+5$ in base $10$ and $52_b = 5b+2$ base $10$. The problem tells us that $5b+2 = 4b+10$, yielding $\boxed{\textbf{(B) }8}$ as our final answer.