2009 AMC 12A Problem 18

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

Write

I_k = 10^{k+2} + 64 = 2^{k+2}5^{k+2} + 2^6

Factor out the smaller power of

2;

the largest count comes at

k = 4,

where

5^6 + 1

supplies one extra factor of

2

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.

Problem 18 in other years