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 $x\leq y$ 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

Other useful links

Pascal’s Identity by argument - Watch and Learn
PRMO-2018, 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 $x\leq y$ 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

Other useful links

Pascal’s Identity by argument - Watch and Learn
PRMO-2018, Problem 11

Leave a Reply Cancel reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

HALL OF FAME BLOG OFFLINE CLASS

CHEENTA ON DEMAND BOSE OLYMPIAD

Cheenta Academy Private Limited |

support@cheenta.com

Menu

Math Olympiad Program