Show that n(2n + 1)(7n + 1) is divisible by 6 for all

[Mellie]

a) Show that n(2n + 1)(7n + 1) is divisible by 6 for all integers n.

b) Find all integers n such that n(2n + 1)(7n + 1) are divisible by 12.

[Guest]

a)

2n+1 is always odd, so any even factors must come from n and/or 7n+1
If n is even then 7n+1 is odd,
and if n is odd then 7n+1 is even,
So n(2n+1)(7n+1) is always even.

Next consider factors of 3.
7n+1 is a multiple of 3 if and only if n+1 is a multiple of 3 (the difference between them is 6n) which in turn holds true if and only if 2n+2 is a multiple of 3 (multiplying by 2 doesn't add or remove factors of 3).
Similarly n is a multiple of 3 if and only if 2n is a multiple of 3.
So one of n, 2n+1, 7n+1 is a multiple of 3 if and only if one of 2n, 2n+1, 2n+2 is a multiple of 3, but this latter statement is obviously true as the list is of 3 consecutive numbers (and multiples of 3 occur once every 3 consecutive numbers).
So n(2n+1)(7n+1) is a multiple of 3

Since n(2n+1)(7n+1) is a multiple of 2 and 3 it must be a multiple of 6.


b)

We already showed n(2n+1)(7n+1) is a multiple of 3, and so it suffices to work out when it is a multiple of 4. As already discussed all even factors must come from n, or 7n+1 (but not both). So n(2n+1)(7n+1) is divisible by 12 when n is a multiple of 4, or when 7n+1 is, the latter condition is equivalent to requiring that n has a remainder of 1 upon division by 4.
(7n+1 is divisible by 4 if and only if -n+1 is (the difference between them is 8n which is a multiple of 4).)

Hope this helped,

R. Baber.