2021 Fall AMC 12B Problems/Problem 6: Difference between revisions
Ehuang0531 (talk | contribs) all 3 solutions are the same |
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~NH14, kingofpineapplz, and Arcticturn | ~NH14, kingofpineapplz, and Arcticturn | ||
== Solution 2 == | |||
We have | |||
<cmath> | |||
\begin{align*} | |||
16,383 & = 2^{14} - 1 \\ | |||
& = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\ | |||
& = 129 \cdot 127 \\ | |||
& = 3 \cdot 43 \cdot 127 . | |||
\end{align*} | |||
</cmath> | |||
Therefore, the greatest prime divisor of 16.383 is 127. | |||
Therefore, the answer is <math>\boxed{\textbf{(C) }10}</math>. | |||
~Steven Chen (www.professorchenedu.com) | |||
==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
Revision as of 21:22, 25 November 2021
- 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 1
We want to find the largest prime factor of
Since
is prime, our answer is
.
~NH14, kingofpineapplz, and Arcticturn
Solution 2
We have
Therefore, the greatest prime divisor of 16.383 is 127.
Therefore, the answer is
.
~Steven Chen (www.professorchenedu.com)
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.