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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<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 2== |
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>. | ||
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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> | ||
| + | |||
| + | ==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 13:41, 14 September 2024
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
