You are given a sequence of $58$ terms; each term has the form $P+n$ where $P$ stands for the product $2 \times 3 \times 5 \times... \times 61$ of all prime numbers less than or equal to $61$, and $n$ takes, successively, the values $2, 3, 4, ...., 59$. let $N$ be the number of primes appearing in this sequence. Then $N$ is:

$ \textbf{(A)}\ 0\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 58 $
1. First, we define \( P \) as the product of all prime numbers less than or equal to 61. Therefore,
   \[
   P = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47 \times 53 \times 59 \times 61
   \]
2. We are given a sequence of 58 terms, each of the form \( P + n \) where \( n \) takes values from 2 to 59. That is, the sequence is:
   \[
   P + 2, P + 3, P + 4, \ldots, P + 59
   \]
3. To determine if any of these terms are prime, we need to consider the properties of \( P \) and \( n \). Since \( P \) is the product of all primes up to 61, it is divisible by each of these primes.
4. For any \( n \) in the range 2 to 59, \( n \) itself must have at least one prime divisor \( p \) such that \( p \leq 59 \). This is because 2 to 59 are all less than or equal to 59, and any integer in this range must have a prime factor less than or equal to itself.
5. Since \( P \) is divisible by all primes up to 61, and \( n \) has a prime divisor \( p \) (where \( p \leq 59 \)), it follows that \( P \) is divisible by \( p \). Therefore, \( P + n \) is also divisible by \( p \).
6. If \( P + n \) is divisible by \( p \) (where \( p \leq 59 \)), then \( P + n \) cannot be a prime number because it has a divisor other than 1 and itself.
7. Consequently, none of the terms \( P + 2, P + 3, \ldots, P + 59 \) can be prime numbers.

Thus, the number of primes \( N \) in this sequence is:
\[
\boxed{0}
\]