Primes and factorials

By the definition of the factorial, all numbers between 2018!+2 and 2018!+2018 (inclusive) are composite. Indeed, the number 2018!+2 is divisible by 2, the number 2018! + 3 is divisible by 3 and so on and so forth. So there is no prime p such that 2018!+1 < p < 2018!+2019.

The number $2018! + x$ (with $1 \leq x \leq 2018$ ) can be factored as $\begin{aligned} 2018! + x & = 1 \cdot 2 \cdots (x-1)\cdot x \cdot (x+1) \cdots 2018 + x\\ & = x \cdot \large(\small 1 \cdot 2 \cdots (x-1) \cdot (x+1)\cdots 2018 + 1\large)\small, \end{aligned}$ showing that it is not prime if $x \geq 2$ .

Alternative

We want prime $\begin{aligned} x! + n \leq p \leq x! +x \phantom{ccc} 1<n \leq x \\ \end{aligned}$ Whenever $n=2k$ where $n\in\mathbb N \leq \left \lfloor \frac{x}{2}\right\rfloor$ . $p\geq\,2k( m +1).$ 2 is the largest even prime which implies $k\,(m+1) = 1\implies m=k \mid 1-k < 0$ shows that no primes exist. Also whenever $n =2k+1$ (odd) . $p \leq \,(2k+ 1)\,(N+1)\implies N +1 =1 \implies N=0$ which is not possible since $N>0$ . So there are no any primes that satisfy the given inequality.

Note: Every prime is an odd number.

The maximun amount of the difference between p and 2018! + 2 and between p and 2018! + 2018 is 2016 (2018-2=2016), which is part of 2018! Hence I can conclude there is no such p exist.

Here’s a not so mathy solution:

First open up your calculator app on your smartphone and attempt to calculate “2018!”

Notice the result is “Error” and thus consider the number to be “very big”

now look at the given problem

• 2018! + 2 <= p <= 2018! + 2018

Subtract 2018! On all sides

• 2 <= p - 2018! <= 2018

Interpret this as

• 2 <= p - number too big for my calculator <= 2018

So this suggests that there may be some prime number p such that the result of p - 2018! Is between 2 & 2018.

Just choose “no”, because the numbers are hella big. Again, not a very mathy solution. But it worked, lol.

If $p = 2018! + n$ where $n \in \{2, 3, \cdots, 2018\}$ , then $n|p$ . Hence, $p$ is composite.

The Prime Number Thoerem formalizes the intuitive idea that primes become less common as they become larger by precisely quantifying the rate at which this occurs. Using this knowledge one can calculate the probability of a number being prime given the size of $N$

Since 2018! and 2018! + 2018 both have 5795 digits

$\log{10^{5795}} = 13343.48$

Means 1 in 13343 numbers is a prime on that scale of digits. Therefore by pure chance the probability that p is prime is 0.000075% Hence you have a high probability of being right by simply saying no.

2018! divides all numbers n for 2 < n < 2018. Thus 2018! + 2 to 2018! + 2018 are composite. The answer is no.

All the numbers from $2018!+2$ to $2018!+2018$ are composite. $2018!+2$ is divisible by $2$ , $2018!+3$ is divisible by 3 . We can say $n! + k$ is divisible by $k$ for $n \ge 2, 2 \le k \le n$ . So there is no such prime. So the answer is $\boxed{NO}$

Let's suppose $p$ exists, then $p=2018!+n$ for some natural number $n$ . As $2\leq n \leq 2018$ , we know that $2018 \mod n =0$ , and then: $p=2018!+n=n(1+\frac{2018!}{n})$ And, therefore, $p$ is not prime, which is a contradiction. In conclusion, there is not such prime number $p$

2018! = 2018 2017 2016 ... 3 2 1 We're given with 2018! + 2 <= p <= 2018! + 2018. Note that if we substitute 2018! with respect to the LHS, the range becomes: 2018 2017 2016 2015 2014 ... 3 2 1 + 2 <= p <= 2018 2017 2016 2015 ... 3 2*1 = 2018.

Here, we generalize the range as n! + 2 <= p <= n! + n, for which p = n! + k, where 2 <= k <= n. By factorization, there is no prime number p that can be found within the range.

Let $k$ be a positive integer $\implies (k + 1)! + j$ where $(2 \leq j \leq k + 1)$ is a list of $k$ consecutive composite integers.

Let $k$ be a positive integer and $(2 \leq j \leq k + 1)$ .

$(k + 1)! + j = j((k + 1) * k * * * (j + 1) * (j - 1) * * * 2 * 1) \implies j|(k + 1)! + j \implies (k + 1)! + j$ is a list of $k$ consecutive composite integers.

For $k = 2017$ we have a list of $2017$ consecutive composite integers. That is, $(2018)! + j$ where $(2 \leq j \leq 2018)$ is a list of $2017$ consecutive composite integers.

Actually, this is a well known theorem that shows there are arbitrarily large gaps between any two primes.

jhgkhjlkjlkjbhgv $LaTeX$

We are looking for a number p=2018!+n. Since this can be written as (2018!/n + 1)*n and we know that with the given conditions, 2018!/n must be integer, the resulting p cannot be a prime number (because we just wrote it as product of two integers, none of which were 1).

i guessed

well if a = b then c = b then why am i doig this is this enough smart to post thanks $LaTeX$ mm latex $\frac{12}{2019!}$

If I divide the 3 members of the inequality by 2018! (inequality does not change direction by dividing by a positive number) I get that there has to be a prime number divided by 2018! between 3 and 2019. That prime number would have to be greater than the right-hand side of the original inequality.

A very clumsy solution but it worked: 2018! is a very very big number. for example, the number of atoms in the observable universe is a billionth of a billionth of billionth...... percent of 2018!. The difference between 2018! + 2 and 2018! + 2018 is 2016, which is a very small gap compared to 2018! And if you look at very big prime numbers, you will realize that they have massive gaps between each other in comparison to 2016. So the probability that a prime number satisfies this is very small.