The L.C.M. and G.C.D. of two numbers are 1530 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the two numbers be $51x$ and $51y$,where $x$ and $y$ are co-prime numbers (i.e.,$gcd(x, y) = 1$).We know that the product of two numbers is eq

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Let the two numbers be $51x$ and $51y$,where $x$ and $y$ are co-prime numbers (i.e.,$gcd(x, y) = 1$).We know that the product of two numbers is equal to the product of their $L.C.M.$ and $G.C.D.$Therefore,$(51x) 	imes (51y) = 51 	imes 1530$.Dividing both sides by $51^2$,we get $x 	imes y = \frac{1530}{51} = 30$.We need to find pairs of co-prime numbers $(x, y)$ such that their product is $30$.The factors of $30$ are $1, 2, 3, 5, 6, 10, 15, 30$.The possible pairs $(x, y)$ such that $x 	imes y = 30$ and $gcd(x, y) = 1$ are:$1) (1, 30)$$2) (2, 15)$$3) (3, 10)$$4) (5, 6)$Since all these four pairs are co-prime,there are $4$ such possible pairs.

The $L.C.M.$ and $G.C.D.$ of two numbers are $1530$ and $51,$ respectively. Find how many such pairs are possible?

Similar Questions

The product of two numbers is $4212$ and their $HCF$ is $9$. Find the possible pair of numbers.

The least number, which when divided by $48, 60, 72, 108$ and $140$ leaves $38, 50, 62, 98$ and $130$ as remainders respectively is

The $HCF$ of $x^{8}-1$ and $x^{4}+2x^{3}-2x-1$ is

Find the greatest number which,on dividing $1657$ and $2037$,leaves remainders $6$ and $5$ respectively.

Two numbers are in the ratio $5:6$. If their $H.C.F.$ is $4$,then their $L.C.M.$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module