Duke
n³ + 11n = n³ + 11n - 12n = n³ - n =
= (n - 1)n(n + 1) ? 0 (mod 6),
and the latter number,
being a product of 3 consecutive naturals is divisible by 6
(at least one of them is even, one of them is multiple of 3)
yasiru89
Or just use induction with the base case n=1
Say holds true for p,
p³ + 11p is divisible by 6, then,
(p+1)³ + 11p + 33 = p³ + 3p² + 14p + 34
= (p³ + 11p) + 3p² + 3p + 34
Now prove the latter bit there is divisible by 6 as a lemma
(its somewhat clumsy but its one of the most general method
for this sort of problem)! (He got his numbers a bit wrong, but the idea is ok.)
Moi
It's fun that this type of problem can be done so many ways:
Instead of checking mod 6, we can check mod 3 and mod 2:
n³ + 11n = n(n² + 11)
mod 2:
n even: product is even
n odd: n³ is odd, 11n is odd, sum is even
mod 3:
0: divisible by n which is divisible by 3
1: n² ~ 1, 1 + 11 = 12, divisible by 3
2: n² ~ 1, 1 + 11 = 12, divisible by 3
Or for the induction step:
(p+1)³ + 11(p+1)
p³ + 3p² + 3p + 1 + 11p + 11
p³ + 11p <--- assumed by induction hypothesis
3p² + 3p + 12
obviously divisible by 3
p even: all terms even
p odd: 3p² and 3p both odd, so sum is even.
This also shows that it works for other numbers than 11:
any number of the form 11 ± 6k would also work
in all the above steps. In other words,
n³ + k*n is divisible by 6 if k ~ 5 (mod 6)
