Question

For $k \gt 0,$ let $I_k = 10\ldots064,$ 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 $N(k)?$

$6$

$7$

$8$

$9$

$10$

Solution:

Write $I_k = 10^{k+2} + 64 = 2^{k+2} \cdot 5^{k+2} + 2^6.$

If $k \lt 4,$ the first term has fewer than $6$ factors of $2,$ so $N(k) = k + 2 \lt 6.$

If $k \gt 4,$ the first term has at least $7$ factors of $2$ while the second has exactly $6,$ so their sum has exactly $6:$ $N(k) = 6.$

If $k = 4,$ then $I_4 = 2^6(5^6 + 1).$ Since $5^6 + 1 = (5^2 + 1)\big((5^2)^2 - 5^2 + 1\big) = 26 \cdot 601,$ it contributes exactly one more factor of $2.$ Thus $N(4) = 7.$

The maximum value is $N(4) = 7.$

Thus, the correct answer is B.

2009 AMC 10A Problem 25

Solution:

Problem 25 in Other Years