Question

Prove that the product of any three consecutive even natural number is divisible by 16 ?

Answer 1

All even numbers are by definition divisible by 2,and we are dealing with consecutive even numbers,so let the numbers be:

$2m,2m+2,2m+4$

Multiplying,the product,$P$is:

$P=2m(2m+2)(2m+4)$

$P=2m(4m2+8m+4m+8)$

$P=8m3+24m2+16m$

Now,pulling out common factor 8m:

$P=8m(m2+3m+2)$

$P=8m(m+1)(m+2)$

Well we know that all such consecutive products will defenitely be divisible by 8,which is a common factor but 16?

Well atleast one of $m,m+1$or$m+2$ must be even, and o could be expressed as $2n$.

Let us assume m is even,for arguments sake,then:

$m=2n$

$\therefore P=8\ast 2n(2n+1)(2n+2)=16n(2n+1)(2n+2)$

Hence product of 3 consecutive even numbers is divisble by 16.