Prove n(n+4)(n+5) is divisible by 6.
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Originally Posted by louis33 Prove n(n+4)(n+5) is divisible by 6. Use induction. If $N(N+4)(N+5)$ is divisible by $6$ then it is plainly obvious so is $(N+1)(N+1+4)(N+1+5)$.
One of $N+4$ and $N+5$ is divisible by 2 (one is odd, the other even). One of: $N$ $N+4 = (N+1)+3$ $N+5 = (N+2)+3$ is divisible by 3 (the other two are not). Any product that is divisible by both 2 and 3 is divisible by 6.
Originally Posted by louis33 Prove n(n+4)(n+5) is divisible by 6. n must be an integer, of course...
Originally Posted by louis33 Prove n(n+4)(n+5) is divisible by 6. $\displaystyle n(n + 4)(n + 5) = $ $\displaystyle n^3 + 9n^2 + 20n = $ $\displaystyle n^3 + 3n^2 + 2n + 6n^2 + 18n =$ $\displaystyle n(n + 1)(n + 2) + 6(n^2 + 3n)$