Difference between revisions of "2015 AMC 10A Problems/Problem 15"
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<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{infinitely many}</math> | <math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{infinitely many}</math> | ||
− | ==Solution== | + | ==Solution 1== |
You can create the equation <math>\frac{x+1}{y+1}=\frac{11x}{10y}</math> | You can create the equation <math>\frac{x+1}{y+1}=\frac{11x}{10y}</math> | ||
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<math>x</math> and <math>y</math> must be positive, so <math>x > 0</math> and <math>y > 0</math>, so <math>x - 10> -10</math> and <math>y + 11 > 11</math>. | <math>x</math> and <math>y</math> must be positive, so <math>x > 0</math> and <math>y > 0</math>, so <math>x - 10> -10</math> and <math>y + 11 > 11</math>. | ||
− | + | Using the factors of 110, we can get the factor pairs: <math>(-1, 110),</math> <math>(-2, 55),</math> and <math>(-5, 22).</math> | |
But we can't stop here because <math>x</math> and <math>y</math> must be relatively prime. | But we can't stop here because <math>x</math> and <math>y</math> must be relatively prime. | ||
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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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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 <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>. | ||
+ | |||
+ | There is only one valid solution so the answer is <math>\boxed{\textbf{(B) }1}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/p7g0hTxE9I8 | ||
+ | |||
+ | ~savannahsolver | ||
==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}} | ||
+ | |||
+ | [[Category: Introductory Algebra Problems]] |
Latest revision as of 14:45, 27 September 2020
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 .
Using the factors of 110, we can get 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
Video Solution
~savannahsolver
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 |
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