Difference between revisions of "2015 AMC 10A Problems/Problem 15"
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==Solution 3== | ==Solution 3== | ||
| − | So from this question, we can get (x+1 | + | So from this question, we can get \(\frac{x+1}{y+1} = \frac{11x}{10y}\). We can transform this equation into \(x + 11 \cdot \left( \frac{x}{y} \right) = 10\). Two numbers are added to get 10, and one of them, \(x\), is a positive and prime integer. So the other number also has to be a positive integer. Therefore, \(11 \cdot \left( \frac{x}{y} \right)\) is a positive integer. The only possibility of this being true is when \(y\) and 11 cancel out, leaving a singular \(x\). So \(y = 11\) and \(x + x = 10\). Therefore, \(y = 11\) and \(x = 5\). |
| − | We can transform this equation into x+11 | ||
| − | |||
~POISONPOISSON | ~POISONPOISSON | ||
Latest revision as of 06:09, 18 December 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
Solution 3
So from this question, we can get \(\frac{x+1}{y+1} = \frac{11x}{10y}\). We can transform this equation into \(x + 11 \cdot \left( \frac{x}{y} \right) = 10\). Two numbers are added to get 10, and one of them, \(x\), is a positive and prime integer. So the other number also has to be a positive integer. Therefore, \(11 \cdot \left( \frac{x}{y} \right)\) is a positive integer. The only possibility of this being true is when \(y\) and 11 cancel out, leaving a singular \(x\). So \(y = 11\) and \(x + x = 10\). Therefore, \(y = 11\) and \(x = 5\).
~POISONPOISSON
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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