5. [4] Call a positive integer $n$ weird if $n$ does not divide $(n-2)$ !. Determine the number of weird numbers between 2 and 100 inclusive.
Answer: 26
We claim that all the weird numbers are all the prime numbers and 4 . Since no numbers between 1 and $p-2$ divide prime $p,(p-2)$ ! will not be divisible by $p$. We also have $2!=2$ not being a multiple of 4 .

Now we show that all other numbers are not weird. If $n=p q$ where $p \neq q$ and $p, q \geq 2$, then since $p$ and $q$ both appear in $1,2, \ldots, n-2$ and are distinct, we have $p q \mid(n-2)!$. This leaves the only case of $n=p^{2}$ for prime $p \geq 3$. In this case, we can note that $p$ and $2 p$ are both less than $p^{2}-2$, so $2 p^{2} \mid(n-2)$ ! and we are similarly done.
Since there are 25 prime numbers not exceeding 100 , there are $25+1=26$ weird numbers.