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I need some help with this problem, can you help me?

Given $k\in\mathbb{Z}^*$ prove $\forall x\in\mathbb{Z}, k\mid x(x+1)(x+2)\cdots(x+k-1)$.

Sorry that one is pretty easy, I made a mistake.

The problem is:

[Update] Given $k\in\mathbb{Z}^*$ prove $\forall x\in\mathbb{Z}, k!\mid x(x+1)(x+2)\cdots(x+k-1)$.

User8976
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JoseA132
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  • The standard way to do this is with binomial coefficients. But also if you carefully do a double induction on $k\ge 0$ and $x\ge 0$ then the argument is fairly straightforward. To extend it to all $x \in \mathbb Z$ you can also use induction or just modular arithmetic mod $k!$. – Erick Wong Feb 12 '17 at 03:37

1 Answers1

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HINT: $x,(x+1),(x+2),\cdots(x+k-1)$ are $k$ consecutive integers.

If $x$ is a multiple of $k$ then you are done. But if not so then by division algorithm there is $q,r \in \Bbb Z$ such that $x = kq +r$ and $0 \leq r <k$.

Now consider $k-r$ and see that $x+k-r$ is among the list and is divisible by $k$.

Consider the Binomial coefficient $$\binom{x+k-1}{x-1}$$ and see your question follows. As It is always an integer and $$\binom{x+k-1}{x-1} = \frac{(x+k-1)!}{(x-1)! \times k! } = \frac{x(x+1)(x+2)\cdots(x+k-1)}{k!}$$

Thus $\forall x\in\mathbb{Z},\ \ k!\mid x(x+1)(x+2)\cdots(x+k-1)$.

User8976
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