Factorial Gaps!

You can divide $100! + n$ by $n$ ( $1 < n \leq 100$ ) this way:

$\frac { 100!+n }{ n } \\ =\quad \frac { 100\cdot 99\cdot ...\cdot n\cdot ...\cdot 2+n }{ n } \\ =\quad 100\cdot 99\cdot ...\cdot (n+1)\cdot (n-1)\cdot ...\cdot 2+1$

Therefore, none of these numbers are primes!

100!+x = x(100!/x + 1). This is not prime for 1<x<=100, since 100! is clearly divisible by x. Therefore, none of the 99 integers are primes.

$n!+m$ is divisible by $m$ if $n\geq{m}$ .

I decided to prove the more general statement that, given a number $c=a!+b$ with $a \geq b$ , $c$ is always composite

Assume we have a number $n$ such that $n=m!+k$ for $m,k \in$ $N$ where $m\geq k>1$ .

By definition of the factorial, $m!$ is divisible by any integers less than or equal to $m$ , including $k$ .

Therefore, $m!=k*r$ for some $r \in$ $N$ . Thus:

$n=k*r+k$

$=k(r+1)$

The only way that this could be prime is if either $k$ or $r+1$ was $1$ and the other was prime. However, by definition, $k>1$ , and $r+1=1 \rightarrow r=0$ , which would mean $k*r=0$ and, therefore, $m!=0$ , which is never true in the real numbers.

$\therefore m!+k$ is not prime Q.E.D.

As a corollary, if we have a number $m!+k$ and some $a|k$ where $1<a\leq m$ , the number can be rewritten as $a*(\frac {m!}{a} + \frac{k}{a})$ and is, therefore, composite. Therefore, if $m!+k$ is prime and $k$ is composite, $k$ 's lower bound is $(m+1)^2$

easy if we consider 100! we can write it in the form of 2 3 4*......100 so for each no we can take something common which means all are non prime

Since $100!$ has many factors like $2, 3, 5, 7, 11, 13, \cdots$ , so no prime numbers in $100! + 2, 100! + 3, 100! + 4, \cdots, 100! + 98, 100! + 99, 100! + 100$ .

Let $n$ be such a number $2 \le n \le 100$

Now consider $100! + n,$

We can write it as multiple of $n$ ,

$\rightarrow n \times (\frac {100!}{n} + 1)$

Here $\frac {100!}{n}$ is intezer as $2 \le n \le 100$ So its a multiple of $n$ and as it is greater than 1. So the number of prime is $\boxed {0}$

Since $100!=100*99*...*2*1$ , if n is in between 1 and 100, it could be factored out. $100!+n=(100*99*....*(n+1)*n*(n-1)*...*2*1)+n=n((100*99*...*(n+1)*(n-1)*....*2*1)+1)$ , which is not prime.

100! + n can always be divided by n for any n from 2 to 100, therefore all of these must be composite numbers.

Purely a swag (scientific guess)... but at such large numbers, the likelihood of a prime number existing within a given range of 100 integers is extremely low.

For any integers $X$ and $Y$ , $((X \cdot Y) + X) mod (X)$ is congruent to $0$ . Therefore, adding any of the integers 1-100 to $100!$ will not make it prime.

Because each of the possible integers from 2 to 100 also appears as a factor in 100!, the resulting sum will be a multiple of said secondary addend.

100! is a multiple of 2, 3, 4, ... , 100, because 100! = 1x2x3x4x...x100, so if we choose any k from {2, 3, 4, ... , 100} then 100! + k is also a multiple of k. The language of congruence fits this question well.

I prefer laymen solutions. So I'll offer one. It's noted that 100! + 1 is absent from the problem. Fair enough.
One may also be aware that large factorials end in multiple zeroes because of multiple 2*5 product of primes. One may simply remove all of + evens quickly, as they will divide by 2.
I'd propose to take any arbitrary prime and play with it. I'm going to chose 23. MJ.
Consider 100! + 23. 100! can be rewritten as 23 * 100 * 99 ... * 2 * 1 or simply 23x. (let x = all the numbers between 1 and 100 be multiplied together except 23) Hence
100! + 23 can be rewritten as. 23(big number + 1) = 23 (x + 1). So it as least a product of 2 whole numbers. Therefore it isn't prime.
Since I chose 23 arbitrarily, I will now add in all solutions for all odds less than 100. Give me a few days.

100! has a common factor with all of 2, 3, 4....., 100 so it is evident that the numbers are divisible by 2, 3, 4....100. Hence none are prime.

We now that m!+k is divisible by k if 0<k<=m, since m=(m)(m-1)...(k)(k-1)(k-2)...(2)(1) . In fact, if we want m consecutive no-prime numbers, we can list them like: (m+1)! + 2, (m+1)! + 3, ...., (m+1)!+m+1

$a=100!+m = m(\frac{100!}{m}+1)$ .

$\frac{100!}{m}$ is integer

$a$ is obviously not prime , because has more than 2 factors , m and . $\frac{100!}{m}+1)$