The largest natural number by which the produ | vedclass

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(C) Let the three consecutive even natural numbers be $2n, 2n+2$,and $2n+4$,where $n$ is a natural number.The product $P = 2n(2n+2)(2n+4) = 8n(n+1)(n+

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(C) Let the three consecutive even natural numbers be $2n, 2n+2$,and $2n+4$,where $n$ is a natural number.The product $P = 2n(2n+2)(2n+4) = 8n(n+1)(n+2)$.Since $n, (n+1)$,and $(n+2)$ are three consecutive integers,their product $n(n+1)(n+2)$ is always divisible by $3! = 6$.Therefore,the product $P$ is divisible by $8 	imes 6 = 48$.Testing with the first three consecutive even numbers: $2 	imes 4 	imes 6 = 48$.Testing with the next set: $4 	imes 6 	imes 8 = 192$,which is $48 	imes 4$.Thus,the largest natural number that always divides the product is $48$.

The largest natural number by which the product of three consecutive even natural numbers is always divisible,is

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