Answer :
We need to show that, based on that assumption, (k+1)(k+2)(k+3) is also
divisible by 6.
(k+1)(k+2)(k+3) = (k+1)(k+2)k + (k+1)(k+2)3 = k(k+1)(k+2) + 3(k+1)(k+2).
By induction hypothesis, the first term is divisible by 6,
and the second term 3(k+1)(k+2) is divisible by 6 because it contains
a factor 3 and one of the two consecutive integers k+1 or k+2 is
even and thus is divisible by 2. Thus it is divisible by both 3 and
2, which means it is divisible by 6. The theorem is proved since
the sum of two multiples of 6 is also a multiple of 6.
Let the numbers be n,n+1,n+2
For a number to be divisible by 6, it must be divisible by 2 and 3 both.
Out of n,n+1,n+2 , one number will be divisible by 2 because if any one of the numbers is odd ,the next number will be even and hence divisible by 2.
Similarly we can say that since the multiples of 3 repeat after intervals of 3, one in 3 consecutive numbers will be divisible by 3.
For a number to be divisible by 6, it must be divisible by 2 and 3 both.
Out of n,n+1,n+2 , one number will be divisible by 2 because if any one of the numbers is odd ,the next number will be even and hence divisible by 2.
Similarly we can say that since the multiples of 3 repeat after intervals of 3, one in 3 consecutive numbers will be divisible by 3.