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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)$xy=x+y+(x,y)$?

• • • Answer: PRMO-2018, Problem 21

Pre College Mathematics

## Try with Hints

At first we have to find out the divisors that satisfy the equation .so we assume that =ak and =bk and try to find out the divisors

Can you now finish the problem ....

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

Therefore   Can you finish the problem... For , then (b-1) divides 2  Therefore a=1+2=3 and a=1+1=2 Now for so     Therefore Therefore 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)$xy=x+y+(x,y)$?

• • • Answer: PRMO-2018, Problem 21

Pre College Mathematics

## Try with Hints

At first we have to find out the divisors that satisfy the equation .so we assume that =ak and =bk and try to find out the divisors

Can you now finish the problem ....

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

Therefore   Can you finish the problem... For , then (b-1) divides 2  Therefore a=1+2=3 and a=1+1=2 Now for so     Therefore Therefore Hence total number of pairs =3

## Subscribe to Cheenta at Youtube

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