Difference between revisions of "2015 AMC 10A Problems/Problem 15"
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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. |
Revision as of 20:03, 8 August 2018
Contents
[hide]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
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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