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1956 AHSME Problems/Problem 31

Revision as of 19:03, 12 February 2021 by Justinlee2017 (talk | contribs) (Solution)
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Solution

The $20^{\text{th}}$ number will be the value of $20_{10}$ in base $4$. Thus, we see \[20_{10} = (1) \cdot 4^2 + (1) \cdot 4^1 + 0 \cdot 4^0\] \[= 110_{4}\]

$\boxed{E}$

~JustinLee2017

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