# 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
3 years, 7 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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

- 3 years, 7 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, $$k+n-c \leq k+n-1$$

$$-c \leq -1$$

$$c \geq 1$$

- 3 years, 7 months ago