If HCF(a, b) = 8,LCM(a, b) = 64 and a b,the | vedclass

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(D) We know that for any two positive integers $a$ and $b$,the product of the numbers is equal to the product of their $HCF$ and $LCM$.$a 	imes b = H

Vedclass · Accepted Answer

$64$

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(D) We know that for any two positive integers $a$ and $b$,the product of the numbers is equal to the product of their $HCF$ and $LCM$.$a 	imes b = HCF(a, b) 	imes LCM(a, b)$Given $HCF(a, b) = 8$ and $LCM(a, b) = 64$,we have:$a 	imes b = 8 	imes 64 = 512$Since $8$ is the $HCF$,both $a$ and $b$ must be multiples of $8$. Let $a = 8x$ and $b = 8y$,where $x$ and $y$ are coprime and $x > y$ (since $a > b$).Substituting these into the product equation:$(8x) 	imes (8y) = 512$$64xy = 512$$xy = 8$Possible pairs $(x, y)$ such that $x$ and $y$ are coprime and $x > y$ are $(8, 1)$.If $x = 8$ and $y = 1$,then $a = 8 	imes 8 = 64$ and $b = 8 	imes 1 = 8$.Since $a > b$,$a = 64$ is the correct value.

If $HCF(a, b) = 8$,$LCM(a, b) = 64$ and $a > b$,then $a = \dots$

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