Difference between revisions of "2006 JBMO Problems/Problem 1"
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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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Thus <math>4n</math> divides <math>(n-1)!</math>. | Thus <math>4n</math> divides <math>(n-1)!</math>. | ||
| + | ~Kris17 | ||
| − | + | == See Also == | |
| − | + | * [[JBMO]] | |
Latest revision as of 15:57, 28 February 2025
Problem
If
is a composite number, then
divides
.
Solution
We shall prove a more stronger result that
divides
for any composite number
which will cover the case of problem statement.
Let
where
.
Let us define set
First let's note that
Now, all multiples of
from
to
Since
we have that
Also, since
we have that
So, we have that
,
in other words,
divides
Now, all multiples of
from
to
Since
we have that
Also, since
so we have that
So, we have that
,
in other words,
divides
Thus
divides
.
~Kris17