1956 AHSME Problems/Problem 31

Problem 31

In our number system the base is ten. If the base were changed to four you would count as follows: $1,2,3,10,11,12,13,20,21,22,23,30,\ldots$ The twentieth number would be:

$\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 38 \qquad\textbf{(C)}\ 44 \qquad\textbf{(D)}\ 104 \qquad\textbf{(E)}\ 110$

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

1956 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 30	Followed by Problem 32
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1956 AHSME Problems/Problem 31

Problem 31

Solution

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