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

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$64$

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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 $HCF(a, b) = 8$,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$The possible pairs of $(x, y)$ such that $x > y$ and $gcd(x, y) = 1$ are $(8, 1)$.If $x = 8$ and $y = 1$,then $a = 8 	imes 8 = 64$ and $b = 8 	imes 1 = 8$.Checking the $LCM$: $LCM(64, 8) = 64$,which matches the given condition.Thus,$a = 64$.

$Q.1.$ $\text{LCM}(7, 49)$	$A. 50$
$Q.2.$ $\text{LCM}(10, 25)$	$B. 49$
	$C. 51$
	$D. 50$

If $HCF(a, b) = 8$,$LCM(a, b) = 64$ and $a > b$,then $a = ?$ (Options: $8, 16, 32, 64$)

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