2021 Fall AMC 12B Problems/Problem 6: Difference between revisions
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<math>\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22</math> | <math>\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22</math> | ||
== Solution | == Solution == | ||
We have | We have | ||
<cmath> | <cmath>\begin{align*} | ||
\begin{align*} | 16383 & = 2^{14} - 1 \\ | ||
& = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\ | & = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\ | ||
& = 129 \cdot 127 \\ | & = 129 \cdot 127 \\ | ||
& = 3 \cdot 43 \cdot 127 . | & = 3 \cdot 43 \cdot 127. | ||
\end{align*} | \end{align*}</cmath> | ||
</cmath> | |||
Therefore, the greatest prime divisor of | Therefore, the greatest prime divisor of <math>16383</math> is <math>127.</math> The sum of its digits is <math>1+2+7=\boxed{\textbf{(C)} \: 10}.</math> | ||
~Steven Chen (www.professorchenedu.com) ~NH14 ~kingofpineapplz ~Arcticturn | |||
~Steven Chen (www.professorchenedu.com) | |||
==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
Revision as of 00:54, 29 January 2022
- The following problem is from both the 2021 Fall AMC 10B #8 and 2021 Fall AMC 12B #6, so both problems redirect to this page.
Problem
The largest prime factor of 
is 
because 
. What is the sum of the digits of the greatest prime number that is a divisor of 
?
Solution
We have
Therefore, the greatest prime divisor of is 
The sum of its digits is 
~Steven Chen (www.professorchenedu.com) ~NH14 ~kingofpineapplz ~Arcticturn
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=1121
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 10 Problems and Solutions | ||
| 2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 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. 
