Prime Trailing Zeros

Each factorial from $0$ to $4$ will add $0$ trailing zeroes to $N$ .

Each factorial from $5$ to $9$ will add $1$ trailing zero to $N$ , since each of these factorials is a multiple of $5^1$ .

Each factorial from $10$ to $14$ will add $2$ trailing zeroes to $N$ , since each of these factorials is a multiple of $5^2$ .

Each factorial from $15$ to $19$ will add $3$ trailing zeroes to $N$ , since each of these factorials is a multiple of $5^3$ .

Each factorial from $20$ to $24$ will add $4$ trailing zeroes to $N$ , since each of these factorials is a multiple of $5^4$ .

Each factorial from $25$ to $29$ will add $6$ trailing zeroes to $N$ , since each of these factorials is a multiple of $5^6$ .

Each factorial from $30$ to $34$ will add $7$ trailing zeroes to $N$ , since each of these factorials is a multiple of $5^7$ .

Each factorial from $35$ to $39$ will add $8$ trailing zeroes to $N$ , since each of these factorials is a multiple of $5^8$ .

Each factorial from $40$ to $44$ will add $9$ trailing zeroes to $N$ , since each of these factorials is a multiple of $5^9$ .

Each factorial from $45$ to $49$ will add $10$ trailing zeroes to $N$ , since each of these factorials is a multiple of $5^{10}$ .

Factorial of $50$ will add $12$ trailing zeroes to $N$ , since this factorial is a multiple of $5^{12}$ .

So, the number of trailing zeroes ( $E$ ) for each $N$ will be:

From these possible values of $E$ , only $\{ 2, 3, 5, 7, 11, 13, 101, 131, 139, 173, 191 \}$ are prime. So, $\color{#3D99F6} \boxed{11}$ values are prime.