I tried this question solving by mathematical induction. But no luck Is there is any easy way to prove that?
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7What is the remainder when $n!$ is divided by any number between $2$ and $n$? – imranfat Aug 01 '20 at 16:37
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2Hint: If $k > 1$ and $k\mid m$ then $k\not \mid m+1$. – fleablood Aug 01 '20 at 16:53
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1Hint: $n!$ IS divisible by EVERY natural number between $2$ and $n$. – fleablood Aug 01 '20 at 16:54
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Hint: If $k|m$ then $k| b + km \iff k|b$. So if $k|m$ and $k\not \mid b$ then $k\not \mid m + b$. Let $m = n!$ and $b=1$. – fleablood Aug 01 '20 at 17:10
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Duplicate $ $ $d\le n\Rightarrow d\mid n!.,$ so $\bmod d!:\ n!\equiv 0,$ so $,\color{#c00}{n!+1\equiv 1},,$ so $,d\mid \color{#c00}{n!+1}\iff d\mid\color{#c00} 1,,$ by divisibility mod reduction $\ \ $ – Bill Dubuque Jun 30 '25 at 15:37
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Hint: If $k|m$ and $k> 1$ then $k\not \mid m+1$.
Pf: Do we need to prove it? If $m = bk$ for some integer $b$ then $m+1 = bk + 1$ and $\frac {m+1}k = b + \frac 1k$ which is not an integer.
And if $2\le k \le n$ then $n! = 1*2*........ * n = \prod_{j:1\le j \le n} j$. So $k$ is between $2.... n$ then it is one of the factors of $n!$.
FORMAL ANSWER (don't read this)
if $2\le k \le n$ then $k$ is a factor of $n! = 1*2*.....*n$. And so $\frac {n!}k$ is an integer. Lets call $\frac {n!}k :=M$. Then $n! = Mk$. So $n!+1 = Mk + 1$ and $\frac {n!+1}k = \frac {Mk+1}k = M + \frac 1k$. But $k > 1$ so $\frac 1k$ is not an integer. So $M + \frac 1k$ is not an integer. So $k\not \mid n! + 1$.
Or a single-line unbelievably short proof:
$n!$ is divisible by all natural number $2$ to $n$ and $1$ is not divisible by any. So $n!+1$ is not divisible by any.
QED
fleablood
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I never want to say don't use induction but... you shouldn't.
But I never want to say don't so:
BASE CASE: $n=2$
$2! + 1=3$. and $2\not \mid 3$.
INDUCTION STEP:
If $k!+1$ is not divisible but any natural number between $2$ and $k$.
Then $(k+1)! + 1 = k!(k+1) + 1= k\cdot k! + k! + 1= k\cdot k! + (k!+1)$. Now any natural number between $2$ and $k$ will divide* $k\cdot k!$ but will not divide $k!+1$ so it will not divide $k\cdot k! + (k!+1)$.
And $k+1$ does divide $k!(k+1)$ but does not divide $1$. SO $k+1$ does not divide $k!(k+1) + 1$.
So no natural number between $2$ and $k+1$ will divide $(k+1)!$. So that is the induction step.
......
SO it can be proven by induction. And it wasn't as hard as I thought it would be. But it wasn't what I'd call easy.
There is a much easier and direct prove that is based on the same idea:
$n!$ is divisible by all natural numbers between $2$ and $n$ and $1$ is not. So $n! + 1$ is not. End of proof. QED.
fleablood
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