Prove that for $n \gt 6$, there is a number $1 \lt k \lt n/2$ that does not divide $n$ My nine year old asked this question at lunch today: Is there a number that is divisible by everything that is half or less than the number?

I immediately answered, "No. I mean, 6. But not for any number bigger than 6."

So I tried to think why that was true, and my first efforts didn't quite work. I did come up with a proof, but it isn't as elegant as I hoped.

How would you prove this?
if $n\geq 7$ and $n$ is divisible by all numbers less then $\frac{n}{2}$, so $n$ is even and we can deduce easily that $\frac{n}{2}(\frac{n}{2}-1)$ divides $n$ (because $\gcd(\frac{n}{2},(\frac{n}{2}-1))=1$) and the most important:
$$\frac{n}{2}(\frac{n}{2}-1)\leq n$$

whence : $n-2=2(\frac{n}{2}-1)\leq 4$ so $n\leq 6$
if n is odd, $f=\frac{n-1}2$ is an integer less than $\frac n2$. $2f=n-1$, so if $f\gt1$ f is not a factor of n. if $f\gt1$ then $n\gt3$, so for all odd $n\gt3$ n there is a number f which is not a factor.
if n is even, $f=\frac{n-2}2$ is an integer less than $\frac n2$. $2f=n-2$, so if $f\gt2$ f is not a factor of n. if $f\gt2$ then $n\gt6$, so for all even $n\gt6$ n there is a number f which is not a factor.