Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 9"

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==Problem==
 
==Problem==
Compute the smallest [[positive integer]] <math>k</math> such that the [[fraction]] <center><p><math>\frac{7k+100}{5k-3}</math></p></center> is [[reducible]].
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Compute the smallest [[positive integer]] <math>k</math> such that the [[fraction]] <center><p><math>\frac{7k+100}{5k-3}</math></p></center> is [[reducible fraction | reducible]].
  
 
==Solution==
 
==Solution==

Latest revision as of 20:58, 12 February 2007

Problem

Compute the smallest positive integer $k$ such that the fraction

$\frac{7k+100}{5k-3}$

is reducible.

Solution

Suppose $p>1$ is a common divisor of $7k+100$ and $5k-3$. Then $p$ also divides $a\cdot (7k+100) + b\cdot (5k-3)$ for integers $a,b$. Putting $a=5$ and $b=-7$ gives $p|521$. Since $521$ is prime and $p>1$, we have $p=521$. Thus $521$ divides $5k-3$, or $5k-3\equiv0\pmod {521}$ or $k\equiv 209\pmod {521}$. Since we are looking for the smallest positive solution, our answer is $209$.