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

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

Question

(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

$64$

Vedclass · Answer

(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,$a = 64$.

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

Similar Questions

When $(5k+1)^2$ is divided by $5$,the remainder is . . . . . . .

The last digit of $2^{m} \cdot 5^{n}$ (where $m, n \in N$) is . . . . . . .

Which of the following is a rational number? $(\sqrt{7}, 3\pi, 0.101000..., 0)$

The product of any four consecutive positive integers is always divisible by:

If $a, b, c$ are distinct prime integers,then the ratio of their $HCF$ to their $LCM$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module