Difference between revisions of "1985 AHSME Problems/Problem 26"

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<math> \gcd(71, n-13) </math>
 
<math> \gcd(71, n-13) </math>
  
Since <math> 71 </math> is prime, <math> n-13 </math> must be a multiple of <math> 71 </math>, which first occurs when <math> n=71+13=84, \boxed{\text{E}} </math>.
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Since <math> 71 </math> is prime, <math> n-13 </math> must be a multiple of <math> 71 </math>, which first occurs when <math> n=71+13=84</math>, <math>\boxed{\text{(E) 84}} </math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=25|num-a=27}}
 
{{AHSME box|year=1985|num-b=25|num-a=27}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:07, 9 May 2020

Problem

Find the least positive integer $n$ for which $\frac{n-13}{5n+6}$ is a non-zero reducible fraction.

$\mathrm{(A)\ } 45 \qquad \mathrm{(B) \ }68 \qquad \mathrm{(C) \  } 155 \qquad \mathrm{(D) \  } 226 \qquad \mathrm{(E) \  }\text84$

Solution

For the fraction to be reducible, the greatest common factor of the numerator and the denominator must be greater than $1$. By the Euclidean algorithm,

$\gcd(5n+6, n-13)$

$\gcd(5n+6-5(n-13), n-13)$

$\gcd(71, n-13)$

Since $71$ is prime, $n-13$ must be a multiple of $71$, which first occurs when $n=71+13=84$, $\boxed{\text{(E) 84}}$.

See Also

1985 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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