Difference between revisions of "2011 AMC 12B Problems/Problem 15"

(See also)
Line 1: Line 1:
 +
==Problem 15==
  
 +
How many positive two-digits integers are factors of <math>2^{24}-1</math>?
 +
 +
<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14</math>
 +
 +
 +
== Solution ==
 +
 +
From repeated application of difference of squares:
 +
 +
<math>2^{24}-1 = (2^{12} + 1)(2^{6} + 1)(2^{3} + 1)(2^{3} - 1)</math>
 +
 
 +
<math>2^{24}-1 = (2^{12} + 1) * 65 * 9 * 7</math>
 +
 +
<math>2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7</math>
 +
 +
Applying sum of cubes:
 +
 +
<math>2^{12}+1 = (2^4 + 1)(2^8 - 2^4 + 1) </math>
 +
 +
<math>2^{12}+1 = 17 * 241</math>
 +
 +
 +
A quick check shows <math>241</math> is prime.  Thus, the only factors to be concerned about are <math>3^2 * 5 * 7 * 13 * 17</math>, since multiplying by <math>241</math> will make any factor too large.
 +
 +
Multiply <math>17</math> by <math>3</math> or <math>5</math> will give a two digit factor; <math>17</math> itself will also work.  The next smallest factor, <math>7</math>, gives a three digit number.  Thus, there are <math>3</math> factors which are multiples of <math>17</math>.
 +
 +
Multiply <math>13</math> by <math>3, 5</math> or <math>7</math> will also give a two digit factor, as well as <math>13</math> itself.  Higher numbers will not work, giving an additional <math>4</math> factors.
 +
 +
Multiply <math>7</math> by <math>3, 5, </math> or  <math> 3^2</math> for a two digit factor.  There are no mare factors to check, as all factors which include <math>13</math> are already counted.  Thus, there are an additional <math>3</math> factors.
 +
 +
Multiply <math>5</math> by <math>3</math> or <math>3^2</math> for a two digit factor.  All higher factors have been counted already, so there are <math>2</math> more factors.
 +
 +
Thus, the total number of factors is <math>3 + 4 + 3 + 2 = \boxed{12    \textbf{ (D)}}</math>
 +
 +
== See also ==
 +
{{AMC12 box|year=2011|ab=B|num-b=14|num-a=16}}
 +
 +
[[Category:Introductory Number Theory Problems]]
 +
{{MAA Notice}}

Revision as of 18:26, 5 February 2018

Problem 15

How many positive two-digits integers are factors of $2^{24}-1$?

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14$


Solution

From repeated application of difference of squares:

$2^{24}-1 = (2^{12} + 1)(2^{6} + 1)(2^{3} + 1)(2^{3} - 1)$

$2^{24}-1 = (2^{12} + 1) * 65 * 9 * 7$

$2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7$

Applying sum of cubes:

$2^{12}+1 = (2^4 + 1)(2^8 - 2^4 + 1)$

$2^{12}+1 = 17 * 241$


A quick check shows $241$ is prime. Thus, the only factors to be concerned about are $3^2 * 5 * 7 * 13 * 17$, since multiplying by $241$ will make any factor too large.

Multiply $17$ by $3$ or $5$ will give a two digit factor; $17$ itself will also work. The next smallest factor, $7$, gives a three digit number. Thus, there are $3$ factors which are multiples of $17$.

Multiply $13$ by $3, 5$ or $7$ will also give a two digit factor, as well as $13$ itself. Higher numbers will not work, giving an additional $4$ factors.

Multiply $7$ by $3, 5,$ or $3^2$ for a two digit factor. There are no mare factors to check, as all factors which include $13$ are already counted. Thus, there are an additional $3$ factors.

Multiply $5$ by $3$ or $3^2$ for a two digit factor. All higher factors have been counted already, so there are $2$ more factors.

Thus, the total number of factors is $3 + 4 + 3 + 2 = \boxed{12    \textbf{ (D)}}$

See also

2011 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png