Difference between revisions of "1960 AHSME Problems/Problem 16"

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==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1960|num-b=15|num-a=17}}
 
{{AHSME 40p box|year=1960|num-b=15|num-a=17}}
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[[Category:Introductory Number Theory Problems]]

Latest revision as of 19:02, 17 May 2018

Problem

In the numeration system with base $5$, counting is as follows: $1, 2, 3, 4, 10, 11, 12, 13, 14, 20,\ldots$. The number whose description in the decimal system is $69$, when described in the base $5$ system, is a number with:

$\textbf{(A)}\ \text{two consecutive digits} \qquad\textbf{(B)}\ \text{two non-consecutive digits} \qquad \\ \textbf{(C)}\ \text{three consecutive digits} \qquad\textbf{(D)}\ \text{three non-consecutive digits} \qquad \\ \textbf{(E)}\ \text{four digits}$

Solution

Since $25<69<125$, divide $69$ by $25$. The quotient is $2$ and the remainder is $19$, so rewrite the number as \[69 = 2 \cdot 25 + 19\] Similarly, dividing $19$ by $5$ results in quotient of $3$ and remainder of $4$, so rewrite the number as \[69 = 2 \cdot 25 + 3 \cdot 5 + 4 \cdot 1\] Thus, the number in base $5$ can be written as $234_5$, so the answer is $\boxed{\textbf{(C)}}$

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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