Product of $N$ Consecutive Integers Divided by $N$

Quite simply,

If you have $n$ consecutive numbers, one of them must be a multiple of $n$ (as multiples of $n$ occur every $n$ integers)

The product of any $N$ consecutive integers not only divisible by $N$ , but also divisible by $N!$ .

Proof: Let the $N$ positive consecutive integers as $k+1,k+2,…,k+(N-1),k+N$ . Consider the integer $\binom{k+N}{k}$ , which is a $\large integer$ . $\begin{aligned} & \binom{k+N}{k} \\ =& \dfrac{(k+N)!}{k! \times N!} \\ =& \dfrac{(k+1) \times (k+2) \times … \times (k+(N-1)) \times (k+N)}{N!} \end{aligned}$ This imply that $\dfrac{(k+1) \times (k+2) \times … \times (k+(N-1)) \times (k+N)}{N!}$ is also an integer, which means that $N! \space | \space [(k+1) \times (k+2) \times … \times (k+(N-1)) \times (k+N)]$

For product of $N$ consecutive negative integers, we just need to take out the negative signs of them, and we will get a number which is product of $N$ consecutive positive integers with negative sign or maybe not. As the negative sign doesn't affect the divisibility, we can conclude that it must me divisible by $N!$ . If the $N$ consecutive number include positive integers and negative integers, then it must include $0$ , so the product is equal to $0$ , which is divisible by any $N!$ .

We must recognize first that─

The product of $N$ consecutive integers is divisible by $N$ if and only if at least one of $N$ consecutive integers is divisible by $N$ .

• Let $a+1, a+2, \ldots \ldots \ldots, a+N$ be $N$ consecutive integers with $a$ is an integer.

• Let's assume that none of $(a+k)$ with $(1\leq k \leq N)$ is divisible by $N$ . So, there remain $(N-1)$ possible remainders, namely, $1, 2, 3, \dots \ldots \ldots, N-1$ , for $N$ integers $(a+k)$ when divided by $N$ .

• $(N-1)$ possible remainders for $N$ integers. So, by the Pigeonhole Principle , at least two of numbers $(a+k)$ will share the same remainder when divided by $N$ .

• Let's assume that $(a+i)$ and $(a+j)$ give same remainder when divided by $N$ , with $1 \leq i < j \leq N$ . So, $(a+j)-(a+i)=(j-i)$ is divisible by $N$ . But as $1 \leq i < j \leq N$ , so, $1\leq (j-i) < N$ . So, $j-i$ cannot be divisible by $N$ . Therefore, assuming none of $N$ consecutive integers is divisible by $N$ leads to a contradiction.

So, $\boxed{\text{YES}}$ , The product of $N$ consecutive integers is divisible by $N$ .

If you're clueless start with simpler problem. Why should product of any two consecutive positive integers be divisible by 2?

In the product of $N$ consecutive integers, one of them must have $N$ as its factor.

Hence the product of any $N$ consecutive integers is always divisible by $N$ .

11x12x13x14x15 is the same as 11x12x13x14x3x5, which is divisible by 5

Since 3 has its multiple after every 3 number So if you choose any 3 consecutive numbers , then there will definitely exist one multiple of 3.

And the same happens with every no. other than 3. Hence it's true

无论如何每行个数等于因数必然排列中存在它的倍数

Since one factor is $15$ (and $15$ is divisible by $5$ ), the product is divisible by $5$ .

This does not hold if zero, which is an integer , is included in the series.

0 x 1 x 2 = 2, which is not divisible by 3.

0 x 1 x 2 x 3 = 6, which is not divisible by 4, etc