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# Largest Common Divisor | PRMO-2014 | Problem 11 Try this beautiful problem from Algebra based on Largest Common Divisor .

## Largest Common Divisor | PRMO | Problem 11

For natural numbers x$x$ and y$y$,let (x,y)$(x,y)$ denote the largest common divisor of x$x$ and y$y$. How many pairs of natural numbers x$x$ and y$y$ with xy$x\leq y$ satisfy the equation xy=x+y+(x,y)$xy=x+y+(x,y)$?

• $1$
• $3$
• $4$

Answer:$3$

PRMO-2018, Problem 21

Pre College Mathematics

## Try with Hints

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

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Try this beautiful problem from Algebra based on Largest Common Divisor .

## Largest Common Divisor | PRMO | Problem 11

For natural numbers x$x$ and y$y$,let (x,y)$(x,y)$ denote the largest common divisor of x$x$ and y$y$. How many pairs of natural numbers x$x$ and y$y$ with xy$x\leq y$ satisfy the equation xy=x+y+(x,y)$xy=x+y+(x,y)$?

• $1$
• $3$
• $4$

Answer:$3$

PRMO-2018, Problem 21

Pre College Mathematics

## Try with Hints

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

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