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

$\overline{abcabc}\  = \overline{abc}\* 1001$ .


~GEOMETRY-WIZARD.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png