Mock AIME 4 2006-2007 Problems/Problem 9

Problem

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

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

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 $209$ .

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Mock AIME 4 2006-2007 Problems/Problem 9

Problem

Solution