1951 AHSME Problems/Problem 19

Problem

A six place number is formed by repeating a three place number; for example, $256256$ or $678678$ , etc. Any number of this form is always exactly divisible by:

$\textbf{(A)}\ 7 \text{ only} \qquad\textbf{(B)}\ 11 \text{ only} \qquad\textbf{(C)}\ 13 \text{ only} \qquad\textbf{(D)}\ 101 \qquad\textbf{(E)}\ 1001$

Solution

We can express any of these types of numbers in the form $\overline{abc}\times 1001$ , where $\overline{abc}$ is a 3-digit number. Therefore, the answer is $\textbf{(E)}\ 1001$ .

Solution 2

$\overline{abcabc}$ =$ 1001 $\overline{abc}\ . So the answer is$ \textbf{(E)}\ 1001$

~GEOMETRY-WIZARD.

1951 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1951 AHSME Problems/Problem 19

Contents

Problem

Solution

Solution 2

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