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# The product of n consectutive natural numbers is divisible by n factorial.

Show that for non-zero natural numbers k and n. $$\left \{k,n \right \}\subset\left \{ \mathbb{N}\setminus \left \{ 0 \right \} \right \}$$

$$k\cdot (k+1)\cdot (k+2)\cdot (k+3)\cdot ...\cdot(k+n-1)$$ is divisible by $$n!$$

Note by Jack Han
2 years, 11 months ago

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Let $$c\equiv k (mod\quad { n })$$ with $$0\le c<n$$ then $$k+n-c$$ is a factor in the product. Since $$c\equiv k(mod\quad { n })$$, then $$(k+n-c)(mod\quad n)\equiv (k-c) (mod\quad n)\equiv 0$$ and since one of the factors is divisible by $$n$$ the product must be as well

- 2 years, 11 months ago

$$k+n-c$$ is factor if $$k$$ is not a factor.

$$1\leq c <n$$

If $$k$$ is divisible by $$n$$ then there is nothing left to prove.

Edit: Also, $$k+n-c \leq k+n-1$$

$$-c \leq -1$$

$$c \geq 1$$

- 2 years, 11 months ago