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

(Solution 2)
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== Solution 2 ==  
 
== Solution 2 ==  
<math>\overline{abcabc}\$ = </math>1001<math> </math>\overline{abc}\$. So the answer is $\$textbf{(E)}\
+
<math>\overline{abcabc}\$ = </math>1001<math> </math>\overline{abc}\$ . So the answer is $\$textbf{(E)}\$
  
 
~GEOMETRY-WIZARD.
 
~GEOMETRY-WIZARD.
 +
 
== See Also ==
 
== See Also ==
 
{{AHSME 50p box|year=1951|num-b=18|num-a=20}}  
 
{{AHSME 50p box|year=1951|num-b=18|num-a=20}}  

Revision as of 06:45, 31 December 2023

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$$ (Error compiling LaTeX. Unknown error_msg)\overline{abc}$ . So the answer is $$textbf{(E)}$

~GEOMETRY-WIZARD.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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