2011 AMC 12B Problems/Problem 15: Difference between revisions

← Older edit Newer edit →

Revision as of 19:54, 29 May 2023

Problem 15

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

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

~ pi_is_3.14

Solution

Repeating difference of squares:

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

$2^{24}-1=(2^{12}+1)\cdot65\cdot9\cdot7$

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

The sum of cubes formula gives us:

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

$2^{12}+1 = 17\cdot241$

A quick check shows $241$ is prime. Thus, the only factors to be concerned about are $3^2\cdot5\cdot7\cdot13\cdot17$ , since multiplying by $241$ will make any factor too large.

Multiplying $17$ 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 that are multiples of $17$ .

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

Multiply $7$ by $3$ , $5$ , or $3^2$ for a two-digit factor. There are no more 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 $3+4+3+2=\boxed{\textbf{(D) }12}$

Video Solution by OmegaLearn

https://youtu.be/mgEZOXgIZXs?t=770

Video Solution by WhyMath

https://youtu.be/5f4yNbRtDOA

~savannahsolver

2011 AMC 12B (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 4: / Line 4: @@
 <math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14</math>
-== Video Solution by OmegaLearn ==
-https://youtu.be/mgEZOXgIZXs?t=770
 ~ pi_is_3.14
@@ Line 12: / Line 9: @@
 == Solution ==
-From repeated application of difference of squares:
+Repeating [[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)(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)\cdot65\cdot9\cdot7</math>
 <math>2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7</math>
-Applying sum of cubes:
+The sum of cubes formula gives us:
+<math>2^{12}+1=(2^4+1)(2^8-2^4+1)</math>
-<math>2^{12}+1 = (2^4 + 1)(2^8 - 2^4 + 1) </math>
+<math>2^{12}+1 = 17\cdot241</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\cdot5\cdot7\cdot13\cdot17</math>, since multiplying by <math>241</math> will make any factor too large.
-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.
+Multiplying <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 that are multiples of <math>17</math>.
-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>.
+Multiplying <math>13</math> by <math>3</math>, <math>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 <math>4</math> additional factors.
-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</math>, <math>5</math>, or  <math>3^2</math> for a two-digit factor. There are no more 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>7</math> by <math>3, 5, </math> or  <math> 3^2</math> for a two digit factor.  There are no more 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.
-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{\textbf{(D) }12}</math>
-Thus, the total number of factors is <math>3 + 4 + 3 + 2 = \boxed{12    \textbf{ (D)}}</math>
+== Video Solution by OmegaLearn ==
+https://youtu.be/mgEZOXgIZXs?t=770
 ==Video Solution by WhyMath==
 https://youtu.be/5f4yNbRtDOA

Page

Toolbox

Search