Question

prove that the product of n successive integers is always divisible by n!

Answer 1

You could use a nice induction proof here.

Answer 2

I would use combinatorics for n in naturals.

Answer 3

Given 1 number, clearly it must be divisible by 1. Given 2 consecutive integers, if you consider their residues modulo 2, the residue classes are {0,1}, so one of them must be divisible by 2. Furthermore, given n consecutive integers, if you consider their residues modulo n, the residue classes are {0,1,2,...,n-1}, so one of them must be divisible by n. Since n is greater than or equal to all numbers {1,2,3...,n}, then clearly n consecutive integers have at least one member that are divisible by each number {1,2,3, ... n}. Thus, their product is divisible by the product 1*2*3*...*n, which is just n!.

Answer 4

For negative numbers there would be two cases:

#1: The product of only negative numbers. This is equal to \( \pm \) the product of the corresponding positive numbers, and so it is divisible by \( n! \).

#2: the product of numbers, some of which are negative, and some of which are positive. But then one of the numbers must be zero, so the product is zero, and zero is divisible by \(n!\)

Hence QED.

Answer 5

And the sum, what Luis Rivera has started is,The sum of \(n\) (odd) consecutive integer is divisible by \(n\) (why?)