Difference between revisions of "2015 AMC 10A Problems/Problem 15"
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<math>(-5, 22)</math> gives <math>x = 5</math> and <math>y = 11</math>. This does work. | <math>(-5, 22)</math> gives <math>x = 5</math> and <math>y = 11</math>. This does work. | ||
− | 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> |
==Solution== | ==Solution== | ||
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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>. | ||
− | Observe that <math>x+1 > \frac{11}{10}\cdot x</math> so <math>x</math> is at most 9. | + | Observe that <math>x+1 > \frac{11}{10}\cdot x</math> so <math>x</math> is at most <math>9.</math> |
By multiplying by <math>\frac{y+1}{x}</math> and simplifying we can rewrite the condition as <math>y=\frac{11x}{10-x}</math>. Since <math>x</math> and <math>y</math> are integer, this only has solutions for <math>x\in\{5,8,9\}</math>. However, only the first yields a <math>y</math> that is relative prime to <math>x</math>. | By multiplying by <math>\frac{y+1}{x}</math> and simplifying we can rewrite the condition as <math>y=\frac{11x}{10-x}</math>. Since <math>x</math> and <math>y</math> are integer, this only has solutions for <math>x\in\{5,8,9\}</math>. However, only the first yields a <math>y</math> that is relative prime to <math>x</math>. | ||
− | There is only one valid solution so the answer is <math>\boxed{\textbf{(B) }1}</math> | + | There is only one valid solution so the answer is <math>\boxed{\textbf{(B) }1}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|num-b=14|num-a=16}} | {{AMC10 box|year=2015|ab=A|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:00, 10 February 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
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
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.