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Difference between revisions of "2006 JBMO Problems/Problem 1"
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==Problem==
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== Problem ==
  
 
If <math>n>4</math> is a composite number, then <math>2n</math> divides <math>(n-1)!</math>.
 
If <math>n>4</math> is a composite number, then <math>2n</math> divides <math>(n-1)!</math>.
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== Solution ==
 
== Solution ==
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Let us define set <math>S = \{1, 2, 3 ... n-1\}</math>
 
Let us define set <math>S = \{1, 2, 3 ... n-1\}</math>
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<math>Case 1: q > p</math>
 
<math>Case 1: q > p</math>
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Thus <math>4n</math> divides <math>(n-1)!</math>.
 
Thus <math>4n</math> divides <math>(n-1)!</math>.
  
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~Kris17
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== See Also ==
  
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<math>Kris17</math>
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* [[JBMO]]

Latest revision as of 15:57, 28 February 2025

Problem

If $n>4$ is a composite number, then $2n$ divides $(n-1)!$.

Solution

We shall prove a more stronger result that $4n$ divides $(n-1)!$ for any composite number $n>4$ which will cover the case of problem statement.

Let $n = p.q$ where $q \ge p \ge 2$.

Let us define set $S = \{1, 2, 3 ... n-1\}$


$Case 1: q > p$

First let's note that $p, q \in S$

Now, all multiples of $p$ from $p.1$ to $p.(q-1) \in S$

Since $q > p => q > 2 => q-1 \ge 2$ we have that $p.2 \in S$ Also, since $p \ge 2$ we have that $2 \in S$

So, we have that $2, p.2, q \in S$, in other words, $4.p.q$ divides $(n-1)!$


$Case 2: q = p$

Now, all multiples of $p$ from $p.1$ to $p.(p-1) \in S$

Since $p > 2 => p-1 \ge 2$ we have that $p.2 \in S$

Also, since $n = p^2 > 4 => p > 2$ so we have that $2 \in S$

So, we have that $2, p.1, p.2 \in S$, in other words, $4.p^2$ divides $(n-1)!$


Thus $4n$ divides $(n-1)!$.

~Kris17

See Also

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