2009 AMC 12A Problem 18

Note that $I_k = 10^{k+2} + 64$ $= 2^{k+2}5^{k+2} + 2^6.$ For $k \lt 4$ the first term has fewer than $6$ factors of $2,$ so $N(k) \lt 6.$ For $k \gt 4$ the first term is divisible by $2^7$ but the $2^6$ term is not, so $N(k) \lt 7.$ For $k = 4,$ $I_4 = 2^6(5^6 + 1).$ Since $5^6 + 1$ $= (5^2 + 1)\big((5^2)^2 - 5^2 + 1\big)$ $= 26\cdot 601,$ and $26 = 2\cdot 13$ contributes exactly one more factor of $2,$ we get $N(4) = 7.$ So the maximum value is $7.$ Thus, the correct answer is B.