Difference between revisions of "1951 AHSME Problems/Problem 19"

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== Solution 2 ==  
 
== Solution 2 ==  
<math>\overline{abcabc}\  = </math>1001$\overline{abc}\$ .(E)
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<math>\overline{abcabc}\  = \overline{abc}\* 1001</math>  .
  
  

Latest revision as of 22:14, 28 July 2024

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
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