Competition problem — Number theory

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Number theory Difficulty 2.0 Junior Find the answer

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:

Pick one

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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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.