The least perfect cube which is completely di | vedclass

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Question

(C) To find the least perfect cube divisible by $21, 24$ and $27$,we first find their Least Common Multiple $(LCM)$.Prime factorization of the numbers

Vedclass · Accepted Answer

$74088$

Vedclass · Answer

(C) To find the least perfect cube divisible by $21, 24$ and $27$,we first find their Least Common Multiple $(LCM)$.Prime factorization of the numbers:$21 = 3 	imes 7$$24 = 2^3 	imes 3$$27 = 3^3$$LCM$ = $2^3 	imes 3^3 	imes 7 = 8 	imes 27 	imes 7 = 1512$.For a number to be a perfect cube,the exponent of each prime factor in its prime factorization must be a multiple of $3$.In the $LCM$ $(2^3 	imes 3^3 	imes 7^1)$,the prime factors $2$ and $3$ already have exponents that are multiples of $3$. However,the prime factor $7$ has an exponent of $1$.To make it a perfect cube,we must multiply the $LCM$ by $7^2$ (which is $49$) so that the exponent of $7$ becomes $3$.Required number = $2^3 	imes 3^3 	imes 7^3 = 1512 	imes 49 = 74088$.

The least perfect cube which is completely divisible by $21, 24$ and $27$ is

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