<s>[INST] Divisibility set of problems It may seem as a begginer question, but i was wondering if there exist a certain method to solve problems like the following :
$$1) \text{ if } n ∈ \Bbb N \\ n(n^2+5)\,⋮\,6$$
$$2)  \text{ if } n ∈ \Bbb N \\n^4+6n^3+11n^2+6n\,⋮\,24$$
$$3)\text{ if } n=2k \text{ then } \frac{n}{12}+\frac{n^2}{8}+\frac{n^3}{24}∈ \Bbb Z $$
NOTE:I havent dont this types of problems in a long time and it would be useful to see a detailed solution.Thanks!a
 [/INST] Another possible method is to write these numbers in terms of binomial coefficients which are always integers :
1) $$\frac{n(n^2+5)}{6}=\frac{n^3+5n}{6}=n+\frac{n^3-n}{6}=n+\frac{n(n+1)(n-1)}{3!}=n+\binom{n+1}{3}$$ which is an integer so $6 \mid n(n^2+5n)$ 
2)  $$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$$ so $$\frac{n^4+6n^3+11n^2+6n}{24}=\frac{n(n+1)(n+2)(n+3)}{4!}=\binom{n+3}{4}$$ which is an integer .
3) $$\frac{k}{6}+\frac{k^2}{2}+\frac{k^3}{3}=\frac{2k^3+3k^2+k}{6}=\frac{k(k+1)(2k+1)}{6}$$
This can be seen either by this method :
This number is actually :
$$2\binom{k+1}{3}+\binom{k+1}{2}$$ 
Or you can do this in an awesome way noticing it's actually :
$$1^2+2^2+3^2+\ldots+k^2$$ (if you know this formula ) .
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