jmc

The number $I_k$ can be written as $10^{k+2}+64=5^{k+2}\cdot 2^{k+2}+2^6$. For $k\in \{1,2,3\}$ we have $I_k=2^{k+2}\left(5^{k+2}+2^{4-k}\right)$. The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have $N(k)=k+2\le 5$. For $k>4$ we have $I_k=2^6\left(5^{k+2}\cdot 2^{k-4}+1\right)$. For $k>4$ the value in the parentheses is odd, hence $N(k)=6$. This leaves the case $k=4$. We have $I_4=2^6\left(5^6+1\right)$. The value $5^6+1$ is obviously even. And as $5\equiv 1(\mathrm{mod}4)$, we have $5^6\equiv 1(\mathrm{mod}4)$, and therefore $5^6+1\equiv 2(\mathrm{mod}4)$. Hence the largest power of $2$ that divides $5^6+1$ is $2^1$, and this gives us the desired maximum of the function $N$: $N(4)=\boxed{7}$.