Difference between revisions of "2007 AMC 12B Problems/Problem 24"
Chickendude (talk | contribs) (New page: ==Problem 24== How many pairs of positive integers <math>(a,b)</math> are there such that <math>gcd(a,b)=1</math> and <cmath>\frac{a}{b}+\frac{14b}{9a}</cmath> is an integer? <math>\mathr...) |
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<math>\mathrm {(A)} 4</math> <math>\mathrm {(B)} 6</math> <math>\mathrm {(C)} 9</math> <math>\mathrm {(D)} 12</math> <math>\mathrm {(E)} \text{infinitely many}</math> | <math>\mathrm {(A)} 4</math> <math>\mathrm {(B)} 6</math> <math>\mathrm {(C)} 9</math> <math>\mathrm {(D)} 12</math> <math>\mathrm {(E)} \text{infinitely many}</math> | ||
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==Solution== | ==Solution== |
Revision as of 13:55, 20 February 2008
Problem 24
How many pairs of positive integers are there such that and is an integer?
Solution
Combining the fraction, must be an integer.
Since the denominator contains a factor of ,
Rewriting as for some positive integer , we can rewrite the fraction
Since the denominator now contains a factor of ,
Rewriting as for some positive integer , we can rewrite the fraction again as
Since the denominator contains ,
Now, returning to and
and
Since , must be . This yields four possible pairs , , ,