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 $n!$

#NumberTheory #NaturalNumbers #Consecutivenumbers

Easy Math Editor

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`2^{34}`	$2^{34}$
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Comments

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

Sean Sullivan - 6 years, 3 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, $k+n-c \leq k+n-1$

$-c \leq -1$

$c \geq 1$

Roman Frago - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$