Lately, while I was practicing for my JEE examination I discovered an amusing fact. I would like to share it with the **Brilliant Community** as a **hypothesis.**

Given a number \(p\), any number \(N\) of the form:

**1.** \(n(n-1)(n+1)(n-2)(n+2)\dots (n-\frac{p-1}{2})(n+\frac{p-1}{2})\) is always divisible by \(p\) **if \(p\) is odd.** \(\left(n\in \text{natural numbers}\right)\)

**2.** \(n(n-1)(n+1)(n+2)(n-2)\dots (n -\frac{p}{2})\) is always divisible by \(p\) **if \(p\) is even.** \(\left(n\in \text{natural numbers}\right)\)

We know that any natural number \(N\)(if want to write it with respect to \(p\)) can be of only the following forms:\(\{kp, kp+1, kp+2,\dots kp+p-1\}\) where \(k\) is a whole number. Now I tried to write an expression for \(N\) such that, no matter what form of number from the above set we put in that expression for \(N\), it is always divisible by \(p\) and that's how I got to those two expressions written above!

This hypothesis has helped me solve quite a number of problems in a sweet and simple fashion. It has at many places helped me reduce certain calculations. It certainly helps you solve problems where \(N\) is given to you as a function of \(n\) and you are asked to find the divisibilty of \(N\) by a certain number. These are a fews ways where this hypothesis can be a bit resourceful!

What I want from the **Brilliant community** is to help me come up with a more refined form of the two expressions if possible and some other amusing observation which when supplemented to this hypothesis can be highly useful. Do comment and share about how you felt about this note. Give as many inputs as possible, this will only help us all grow as a science community.

This was my first note and I will be highly grateful to the community if they provide feedbacks which can help me refine my note writing skills.

Thank you!

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## Comments

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TopNewestYour hypothesis is certainly correct, and a weaker version of the well known result that the product of n consecutive integers is divisible by n! .

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