Largest Common Divisor | PRMO-2014 | Problem 11

At first we have to find out the divisors that satisfy the equation \( xy=x+y+(x,y)\) .so we assume that \(x\)=ak and \(y\)=bk and try to find out the divisors

Can you now finish the problem ....

Let \(x\) =ak and \(y\)=bk, then (x,y)=k and (a,b)=1

Therefore \(xy=x+y+(x,y)\)

\(\Rightarrow abk^2=ka+kb+k\)

\(\Rightarrow kab=a+b+1\)

Can you finish the problem...

\(k=1\Rightarrow ab=a+b+1\Rightarrow a=1=\frac{2}{b-1}\)

For \(a\in \mathbb N\), then (b-1) divides 2

\(\Rightarrow b-1=1,2\)

\(\Rightarrow b=2,3\)

Therefore a=1+2=3 and a=1+1=2

\(\Rightarrow (x,y)=(3,2) or (2,3)\)

Now for \(x=y\Rightarrow(x,y)=x\)

so \(xy=x+y(x<y)\)

\(\Rightarrow x^2=3x\)

\(\Rightarrow x^2 -3x=0\)

\(\Rightarrow x(x-3)=0\)

\(\Rightarrow x=3 or 0\)

Therefore \(x\in \mathbb N\Rightarrow x=3,y=3\)

Therefore \((x,y)=(3,3)\)

Hence total number of pairs =3

At first we have to find out the divisors that satisfy the equation \( xy=x+y+(x,y)\) .so we assume that \(x\)=ak and \(y\)=bk and try to find out the divisors

Can you now finish the problem ....

Let \(x\) =ak and \(y\)=bk, then (x,y)=k and (a,b)=1

Therefore \(xy=x+y+(x,y)\)

\(\Rightarrow abk^2=ka+kb+k\)

\(\Rightarrow kab=a+b+1\)

Can you finish the problem...

\(k=1\Rightarrow ab=a+b+1\Rightarrow a=1=\frac{2}{b-1}\)

For \(a\in \mathbb N\), then (b-1) divides 2

\(\Rightarrow b-1=1,2\)

\(\Rightarrow b=2,3\)

Therefore a=1+2=3 and a=1+1=2

\(\Rightarrow (x,y)=(3,2) or (2,3)\)

Now for \(x=y\Rightarrow(x,y)=x\)

so \(xy=x+y(x<y)\)

\(\Rightarrow x^2=3x\)

\(\Rightarrow x^2 -3x=0\)

\(\Rightarrow x(x-3)=0\)

\(\Rightarrow x=3 or 0\)

Therefore \(x\in \mathbb N\Rightarrow x=3,y=3\)

Therefore \((x,y)=(3,3)\)

Hence total number of pairs =3