Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1985 AHSME Problems/Problem 26"
(Fixed answer choices and improved formatting)
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
Find the least [[positive integer]] <math> n </math> for which <math> \frac{n-13}{5n+6} </math> is a non-zero reducible fraction.
+
Find the least positive integer <math>n</math> for which <math>\frac{n-13}{5n+6}</math> is a non-zero reducible fraction.
  
 
<math> \mathrm{(A)\ } 45 \qquad \mathrm{(B) \ }68 \qquad \mathrm{(C) \  } 155 \qquad \mathrm{(D) \  } 226 \qquad \mathrm{(E) \  }\text{none of these} </math>
 
<math> \mathrm{(A)\ } 45 \qquad \mathrm{(B) \ }68 \qquad \mathrm{(C) \  } 155 \qquad \mathrm{(D) \  } 226 \qquad \mathrm{(E) \  }\text{none of these} </math>
  
 
==Solution==
 
==Solution==
−
For the fraction to be reducible, the greatest common factor of the numerator and the denominator must be greater than <math> 1 </math>. By the [[Euclidean algorithm]],
+
For the fraction to be reducible, the greatest common factor of the numerator and the denominator must be greater than <math>1</math>. Using the [[Euclidean algorithm]], we compute <cmath>\begin{align*}\gcd\left(5n+6,n-13\right) &= \gcd\left(5n+6-5(n-13),n-13\right) \\ &= \gcd\left(71,n-13\right).\end{align*}</cmath>
−
 
+
Since <math>71</math> is prime, it follows that this GCD will be <math>1</math> unless <math>(n-13)</math> is a multiple of <math>71</math>, which first occurs when <math>n = 71+13 = 84</math>, so the answer is <math>\boxed{\text{(E) none of these}}</math>.
−
<math> \gcd(5n+6, n-13) </math>
 
−
 
 
−
<math> \gcd(5n+6-5(n-13), 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>.
 
  
 
==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}}

Latest revision as of 00:44, 20 March 2024

Problem

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

$\mathrm{(A)\ } 45 \qquad \mathrm{(B) \ }68 \qquad \mathrm{(C) \ } 155 \qquad \mathrm{(D) \ } 226 \qquad \mathrm{(E) \ }\text{none of these}$

Solution

For the fraction to be reducible, the greatest common factor of the numerator and the denominator must be greater than $1$. Using the Euclidean algorithm, we compute \begin{align*}\gcd\left(5n+6,n-13\right) &= \gcd\left(5n+6-5(n-13),n-13\right) \\ &= \gcd\left(71,n-13\right).\end{align*} Since $71$ is prime, it follows that this GCD will be $1$ unless $(n-13)$ is a multiple of $71$, which first occurs when $n = 71+13 = 84$, so the answer is $\boxed{\text{(E) none of these}}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
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
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1985_AHSME_Problems/Problem_26&oldid=217176"