Difference between revisions of "1985 AHSME Problems/Problem 26"
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==Problem== | ==Problem== | ||
− | Find the least | + | 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>. | + | 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>. | |
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− | Since <math> 71 </math> is prime, <math> n-13 </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}} |
Latest revision as of 00:44, 20 March 2024
Problem
Find the least positive integer for which is a non-zero reducible fraction.
Solution
For the fraction to be reducible, the greatest common factor of the numerator and the denominator must be greater than . Using the Euclidean algorithm, we compute Since is prime, it follows that this GCD will be unless is a multiple of , which first occurs when , so the answer is .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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All AHSME Problems and Solutions |
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