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

  • 1
  • 3
  • 4

Check the Answer


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

  • 1
  • 3
  • 4

Check the Answer


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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