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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, 8 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

Sean Sullivan - 2 years, 8 months ago

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\(k+n-c\) 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 - 2 years, 8 months ago

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