# Product of 5 Consecutive Integers is Divisible by 120

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## Theorem

Let $a, b, c, d, e \in Z$ be consecutive integers

Then their product $a b c d e$ is divisible by $120$.

## Proof

This theorem requires a proof.In particular: tedious, someone else have a go. Or it's an instance of a general result where the product of $n$ consecutive integers is divisible by $n!$You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{ProofWanted}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Sources

- 1980: David M. Burton:
*Elementary Number Theory*(revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $14$