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> | ||
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
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. 
