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Every integer of the form $(n^{3}-n)(n^{2}-4)$ $(for n = 3,4,....K)$ is

(A) divisible by $6$ but not always divisible by $12$;

(B) divisible by 12 but not always divisible by 24;

(C) divisible by 24 but not always divisible by 120;

(D) divisible by 120 but not always divisible by 720.

Simplifying the equation did not help, and after solving it for $3$ and $4$ we see that it is divisible by everything except $720$.

But how do we find out if this holds true for the entire sequence?

user405925
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2 Answers2

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Hint $$(n^3-n)(n^2-4)=(n-2)(n-1)n(n+1)(n+2).$$ This is a product of five consecutive integers, hence divisible by $5!$

Anurag A
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  • Is the product of five consecutive integers always divisible by $5!$? I understand if there are five consecutive integers in a product, then one must be a multiple of 5 and at least one is multiple of 3, and two or three of them are going to be multiples of 2. – Anna SdTC Feb 04 '17 at 07:10
  • @AnnaSdTC Yes it is a very general result that the product of $k$ consecutive integers is divisible by $k!$. Consider $\binom{n}{k}=\frac{n(n-1)..,(n-k+1)}{k!}$. The left side is an integer, so the right side is also an integer. The numerator is a product of $k$ consecutive integers. Now with proper interpretation of the binomial we can extend the result for negative integers as well. – Anurag A Feb 04 '17 at 07:13
  • That is so neat! – Anna SdTC Feb 04 '17 at 07:14
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D)

$ \forall n \implies 5! \mid P(n) $

$ \forall n \geq 3 \implies (3 \mid n) \oplus (6! \mid P(n)) $

Examples:

$ 5! \mid P(15) $

$ 6! \nmid P(15) $

$ 6! \mid P(13) $

Angel Moreno
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