Difference between revisions of "2015 AMC 10A Problems/Problem 15"
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We found one valid solution so the answer is <math>\boxed{\textbf{(B) }1}</math> | We found one valid solution so the answer is <math>\boxed{\textbf{(B) }1}</math> | ||
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The condition required is <math>\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}</math>. | The condition required is <math>\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}</math>. |
Revision as of 09:32, 4 July 2015
Contents
Problem
Consider the set of all fractions , where and are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by , the value of the fraction is increased by ?
Solution 1
You can create the equation
Cross multiplying and combining like terms gives .
This can be factored into .
and must be positive, so and , so and .
This leaves the factor pairs: and
But we can't stop here because and must be relatively prime.
gives and . and are not relatively prime, so this doesn't work.
gives and . This doesn't work.
gives and . This does work.
We found one valid solution so the answer is
Solution 2
The condition required is .
Observe that so is at most
By multiplying by and simplifying we can rewrite the condition as . Since and are integer, this only has solutions for . However, only the first yields a that is relative prime to .
There is only one valid solution so the answer is
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.