For $ k>0$, let $ I_k\equal{}10\ldots 064$, where there are $ k$ zeros between the $ 1$ and the $ 6$.  Let $ N(k)$ be the number of factors of $ 2$ in the prime factorization of $ I_k$.  What is the maximum value of $ N(k)$?

$ \textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$
1. We start by expressing \( I_k \) in a more manageable form. Given \( I_k = 10\ldots064 \) with \( k \) zeros between the 1 and the 6, we can write:
   \[
   I_k = 10^{k+2} + 64
   \]
   This is because \( 10^{k+2} \) represents the number 1 followed by \( k+2 \) zeros, and adding 64 gives us the desired number.

2. Next, we need to determine the number of factors of 2 in the prime factorization of \( I_k \). We start by factoring \( 64 \):
   \[
   64 = 2^6
   \]
   Therefore, we can rewrite \( I_k \) as:
   \[
   I_k = 10^{k+2} + 2^6
   \]

3. We now consider the factorization of \( 10^{k+2} \). Since \( 10 = 2 \times 5 \), we have:
   \[
   10^{k+2} = (2 \times 5)^{k+2} = 2^{k+2} \times 5^{k+2}
   \]
   Thus, we can express \( I_k \) as:
   \[
   I_k = 2^{k+2} \times 5^{k+2} + 2^6
   \]

4. To find the number of factors of 2 in \( I_k \), we factor out the common power of 2 from both terms:
   \[
   I_k = 2^6 (2^{k-4} \times 5^{k+2} + 1)
   \]
   Here, we have factored out \( 2^6 \) from both terms. The remaining term inside the parentheses is \( 2^{k-4} \times 5^{k+2} + 1 \).

5. We need to determine the number of factors of 2 in \( 2^{k-4} \times 5^{k+2} + 1 \). Notice that:
   \[
   2^{k-4} \times 5^{k+2} + 1
   \]
   For \( k \geq 6 \), \( 2^{k-4} \times 5^{k+2} \) is an even number, and adding 1 to an even number results in an odd number. Therefore, \( 2^{k-4} \times 5^{k+2} + 1 \) has no factors of 2.

6. Consequently, the number of factors of 2 in \( I_k \) is determined solely by the \( 2^6 \) term we factored out:
   \[
   I_k = 2^6 (2^{k-4} \times 5^{k+2} + 1)
   \]
   Since \( 2^{k-4} \times 5^{k+2} + 1 \) is odd, it contributes no additional factors of 2.

7. Therefore, the maximum number of factors of 2 in \( I_k \) is 6, which occurs when \( k \geq 6 \).

The final answer is \( \boxed{6} \).