Factorial factors

Any composite number $n$ (except for squares) will be a factor of $(n-1)!$ because it can be expressed as the product of 2 smaller distinct integers which are present in $(n-1)!$ .

Every square $n$ will be the $\sqrt{n}$ th multiple of $\sqrt{n}$ , so there will be at least $\sqrt{n}-1$ factors of $\sqrt{n}$ in $(n-1)!$ , one for each of the previous multiples of $\sqrt{n}$ . This makes $n$ a factor of $(n-1)!$ , because a square is made up of 2 factors of $\sqrt{n}$ . The only exception to this is when $n=4$ , because there will be only $\sqrt{4}-1=1$ factor of $\sqrt{4}=2$ in $(4-1)!$ , which makes $2^2=4$ one such integer that satisfies the condition in the question.

Any prime number will satisfy the condition in the question because it will be the smallest positive integer that contains all its factors, namely, itself. There are 25 prime numbers less than 100. So, adding 4 to our list, we have $26$ such integers.

The Wilson's Theorem says that n is a prime number only if:

• (n - 1)! ≡ -1 (mod n)

Also, it's possible to proof that whenever n is a composite number and n is different from 4:

• (n-1)! ≡ 0 (mod n)

So we just need to compute how many prime numbers exists between 2 and 100, which is 25 and add 1 cause 4 is an exception.

So the answer is 26.

on one hand : if p is a composite number >4 then (p-1)! is divisible by p. Let's prove that. let p a composite number >4. then p= ab with 1 < a < p and 1 < b < p then a<=p-1 and b<=p-1 - if a is different from b then ab divides (p-1)!=(p-1) x (p-2)x...x b x...x a x...2 x 1 ( if b>a for ex) - if a=b then p=a² ,so a>2, then p-a>a and both a and (p-a) are lower than p, ax(p-a) divides (p-1)! as a divides p-a then a² divides (p-1)!. on an other hand, if p is a prime number by Theorem wilson p doesn't divide (p-1)! the integers n that are 2≤n≤100 and (n−1)! is not divisible by n are the prime numbers between 2 and 100, and 4 so exactly 26 one.

The easiest and quickest way :

The answer is all the prime numbers below $100$ and including $4$

See all prime numbers less than 101 and greater than 1 satisfy this condition and there are 26 prime numbers less than 101 and greater than 1. So the answer is 26

It can easily be seen that number of primes between 2 and 100 would surely in be the answer which is 25.Now let's check composites first is 4. 4 is not divisible by $3!=6$ .Next composite number after 4 is 6 but $5!$ is divisible by 6 and all the composites after that also satisfy the condition.

So the answer is 25+1=26