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1965 AHSME Problems/Problem 15

Revision as of 10:12, 10 June 2016 by Dot22 (talk | contribs) (Created page with "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...")
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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 $\boxed{8}$ as our final answer.

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