# Long Runs with No Primes

In the sequence of all integers, there are arbitrary long runs with no primes.

Yes, indeed. The basic construction constriction starts with factorial. As we know, N! = 1·2·3·4· ... ·N is the product of all integers from 1 through N. By definition, N! is divisible by every integer not exceeding N. Also, in general, a common divisor of any two numbers a and b divides their sum as well.

Thus equipped, we may claim that

N! + 2 is divisible by 2
N! + 3 is divisible by 3
N! + 4 is divisible by 4
N! + 5 is divisible by 5
...
N! + N is divisible by N

It follows that (N - 1) consecutive numbers (N! + 2), (N! + 3), ..., (N! + N) are all composite.

N! grows astronomically fast. 10! = 3628800. 70! is larger than the Googol, which is a 1 followed by 100 zeros. 100! is a number with 158 digits. So we are fortunate to have a short and handy notation for this number.Factorials are indeed used everywhere in Mathematics. Their appearance in the Binomial Theorem and Taylor series would alone justify a special notation for this product. The word "factorial" is in use from 1800 and the notation N! from 1808. Here is a problem that puts to test your understanding of this notation.

## Problem

Find integer m, n, k (all > 1) such that m!n! = k!.

## Solution

Remeber the recursive formula for factorials:

N! = N · (N-1)!?

Well, this suggests a general solution. Choose m arbitrarily and let N = m!, n = N-1, and k = N. Written explicitly, m! × (m! - 1)! = (m!)!.

However, this solution does not cover all the possibilities. For example, 6!7! = 10! is not included in the general formula.

## Reference

1. R. Honsberger, More Mathematical Morsels, MAA, New Math Library, 1991

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