is it a prime

$2019 = 3\cdot673$ , so we have $\begin{array}{llc} 2^{2019} - 1 &= 2^{3\cdot673}-1&~ \\~\\&= (2^{673})^3-1&\color{#20A900}{\text{Notice that this is }x^3-1\text{ form}} \\~\\&=\boxed{(2^{673}-1)(2^{2\cdot673} + 2^{673}+1)}&\color{#20A900}{\text{factored into }(x - 1) (x^2 + x + 1)}\end{array}$ Therefore $2^{2019}-1$ is not a prime number.

We know from here if $2^{n} - 1$ is prime then $n$ must be prime, thus we know from the contrapositive if $n$ is not prime then $2^n - 1$ is not prime. We remember that if the sum of the digits of a number is divisible by three then the original number is divisible by 3 from here and we know that the sum of the digits in $2019$ is $12$ which is divisible by 3, so then we know $2019$ is not prime and that means that $2^{2019} - 1$ is not prime.

The way I solved this was that in base 2, (2**2019)-1 is simply a string of 2019 1s. Because 2019 is divisible by 3, you can partition the binary string into 673 groupings of 3 1s. This means that in base 2, the number can be factored into 111 * 10010010...01001. So it is not prime since it is divisible by 111 in base 2, which is 7 in base 10.

I found that 2^2019-1 is divisible by 7 by listing powers of 2 in modulo 7 to get 2,4,1,2,4,1... From there I found 2^2019 is congruent to 1(mod 7) and so 2^2019-1 is divisible by 7

Since 2019=673 * 3; 2^2019 is divisible by 2^673-1