The product of two numbers is 4212 and their | vedclass

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(B) Let the two numbers be $9a$ and $9b$,where $a$ and $b$ are co-prime numbers.Given that the product of the numbers is $4212$.So,$(9a) 	imes (9b) =

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$36, 117$

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(B) Let the two numbers be $9a$ and $9b$,where $a$ and $b$ are co-prime numbers.Given that the product of the numbers is $4212$.So,$(9a) 	imes (9b) = 4212$.$81ab = 4212$.$ab = \frac{4212}{81} = 52$.Now,we find pairs of co-prime factors of $52$. The factors of $52$ are $(1, 52)$ and $(4, 13)$.Case $1$: If $a=1, b=52$,then the numbers are $9 	imes 1 = 9$ and $9 	imes 52 = 468$.Case $2$: If $a=4, b=13$,then the numbers are $9 	imes 4 = 36$ and $9 	imes 13 = 117$.Comparing with the given options,the pair $(36, 117)$ is correct.

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

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