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. | + | 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== | ||
| − | Suppose <math>p>1</math> is a common divisor of <math>7k+100</math> and <math>5k-3</math>. Then <math>p</math> also divides <math>a\cdot (7k+100) + b\cdot (5k-3)</math> for | + | Suppose <math>p>1</math> is a [[common divisor]] of <math>7k+100</math> and <math>5k-3</math>. Then <math>p</math> also divides <math>a\cdot (7k+100) + b\cdot (5k-3)</math> for [[integer]]s <math>a,b</math>. Putting <math>a=5</math> and <math>b=-7</math> gives <math>p|521</math>. Since <math>521</math> is prime and <math>p>1</math>, we have <math>p=521</math>. Thus <math>521</math> divides <math>5k-3</math>, or <math>5k-3\equiv0\pmod {521}</math> or <math>k\equiv 209\pmod {521}</math>. Since we are looking for the smallest [[positive]] solution, our answer is <math>209</math>. |
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Latest revision as of 19:58, 12 February 2007
Problem
Compute the smallest positive integer 
such that the fraction
is reducible.
Solution
Suppose 
is a common divisor of 
and 
. Then 
also divides 
for integers 
. Putting 
and 
gives 
. Since 
is prime and , we have 
. Thus divides , or 
or 
. Since we are looking for the smallest positive solution, our answer is 
.
