Is $4\underbrace{999 . . . 9}_{224 ({\rm times})}$ prime? Is $4\underbrace{999 . . . 9}_{224 ({\rm times})}$ prime?

I wanted to find smallest prime its sum of digits is $2020$. I started with small primes; the smallest three digits prime its sum of digits is 22 is $499$; four digits is $4999$ with sum of digits 31, five digit is $49999$ with sum of 40.For the sum $2020$ we have:

$2020=224\times 9+4$ and desired number can be of the form  $4\underbrace{999 . . . 9}_{224 ({\rm times})}$ . So this number has at least 225 digits. If it is not prime we have to search for numbers with number of digits more than 225 which of course have digits less than 9 and first digit probably less than 4.   I could not check it with my computer. I have these questions:

1- is  $4\underbrace{999 . . . 9}_{224 ({\rm times})}$  primes?

2- are numbers of the form $499 . . . 99$  always primes? If so what is theoretical reason? If not   what is conditions  for it to be prime? 

Update: the closed form of these numbers is $N=5\times 10^n-1=5(10^n-1)+ 4$, $n ≥ 2$ if n is even we have:

$10^{2k}-1=(10^k-1)(10^k+1)$

Since $[10^n-1, 5, 4]=1$ N can be a prime, but brute force gives a counter result. If n is odd N can be composite.
A computer search finds $4259\mid 5\times10^{224}-1$. I know of no elegant proof of this, just brute force.
is prime
the form $N=5 \times 10^n-1$ is not prime if $n \neq1,7,13,19,25,...,n-1 \neq 6b$