show that profuct of 3 positive integer is divisible by 6
Answers
6×6×6÷6
216÷6
=36
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if x = 3n + 1 ,then x + 2 = 3n + 1 + 2 = 3n + 3
=> x = 3(n + 1) is divisible by 3
if x = 3n + 2, then x + 1 = 3n + 2 + 1
=> 3n + 3 = 3(n + 1)
so, we can say that one of the numbers n, n + 1 and n + 2 is always divisible by 3.
n (n + 1) (n + 2) is divisible by 3.
now,
similarly, when a no. is divisible by 2 remainder abtained is 0 or 1.
therefore,
x = 2r or (2r + 1)
if x = 2r, then x = 2r and (x + 2) then,2r + 2
=> 2(r + 1) are divisible by 2
if x = (2r + 1), then x + 1 = 2r + 1 + 1 = 2(r + 1)is divisible by 2.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 2.
x (x + 1) (x + 2) is divisible by 2.
n (n + 1) (n + 2) is divisible by 2 and 3.
therefore,
n (n + 1) (n + 2) is divisible by 6.