2011 AMC 12B Problem 15

Attempt Problem 15 of the 2011 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12B solutions, or check the answer key.

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15.

How many positive two-digit integers are factors of

2^{24}-1?

4

8

10

12

14

Repeatedly apply the difference of squares to factor

2^{24}-1

The prime factorization is

3^2\cdot5\cdot7\cdot13\cdot17\cdot241;

list two-digit products

Factoring,

\begin{aligned} 2^{24}-1 &=(2^{12}-1)(2^{12}+1) \\ &=(2^6-1)(2^6+1) \\ &\quad {}\cdot(2^4+1)(2^8-2^4+1), \end{aligned}

which equals

63\cdot65\cdot17\cdot241

=3^2\cdot5\cdot7\cdot13\cdot17\cdot241.

Since

241

is a three-digit prime, the two-digit factors come from

3^2\cdot5\cdot7\cdot13\cdot17.

They are

\begin{gathered} 13,15,17,21,35,39,45,51, \\ 63,65,85,91, \end{gathered}

for a total of

12.

Thus, the correct answer is D.

Problem 15 in other years