Set of Fractions | AMC 10A, 2015| Problem No 15

Pre College Mathematics

AMC-10A, 2015 Problem-15

$1$

According to the questation we can write $\frac{x+1}{y+1}=\frac{11 x}{10 y}$

\(\Rightarrow xy +11x-10y=0\)

\(\Rightarrow (x-10)(y-11)=-110\)

Now can you finish the problem?

Here \(x\) and \(y\) must positive, so $x>0$ and $y>0$, so $x-10>-10$ and $y+11>11$

Now we have to find out the factors of \(110\) and find out the possible pairs to fulfill the condition....

Now Can you finish the Problem?

uses the factors of $110$ , we can get the factor pairs: $(-1,110),(-2,55),$ and $(-5,22)$
But we can't stop here because $x$ and $y$ must be relatively prime.
$(-1,110 )$ gives $x=9$ and $y=99.9$ and 99 are not relatively prime, so this doesn't work.
$(-2,55 )$ gives $x=8$ and $y=44$. This doesn't work.
$(-5,22)$ gives $x=5$ and $y=11$. This does work.

Therefore the one solution exist

Pre College Mathematics

AMC-10A, 2015 Problem-15

$1$

According to the questation we can write $\frac{x+1}{y+1}=\frac{11 x}{10 y}$

\(\Rightarrow xy +11x-10y=0\)

\(\Rightarrow (x-10)(y-11)=-110\)

Now can you finish the problem?

Here \(x\) and \(y\) must positive, so $x>0$ and $y>0$, so $x-10>-10$ and $y+11>11$

Now we have to find out the factors of \(110\) and find out the possible pairs to fulfill the condition....

Now Can you finish the Problem?

uses the factors of $110$ , we can get the factor pairs: $(-1,110),(-2,55),$ and $(-5,22)$
But we can't stop here because $x$ and $y$ must be relatively prime.
$(-1,110 )$ gives $x=9$ and $y=99.9$ and 99 are not relatively prime, so this doesn't work.
$(-2,55 )$ gives $x=8$ and $y=44$. This doesn't work.
$(-5,22)$ gives $x=5$ and $y=11$. This does work.

Therefore the one solution exist