Prove that $6|2n^3+3n^2+n$

Solution 1:

If you write this as $2n^3+3n^2+n\equiv 0 \mod 6$ then you only need to check $n=0,1,2,3,4,5$.

Alternatively, write as $\dfrac{2n(2n+1)(2n+2)}{4}$ where the numerator obviously has a multiple of 3, a multiple of 4 and another multiple of 2, so is divisible by 24, meaning the expression is divisible by 6.

Solution 2:

You have $2n^3+3n^2+n=n(n+1)(2n+1)$, and $2\mid n(n+1)$. If $3\mid n(n+1)$, then you're done. Otherwise, $n\not\equiv 0\pmod{3}$ and $n\not\equiv -1\pmod{3}$, so $n\equiv 1\pmod{3}$. Then $2n+1\equiv 3\equiv 0\pmod{3}$, so $3\mid 2n+1$ and you get the result.