Difference between revisions of "2022 AMC 10B Problems/Problem 6"
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Revision as of 23:15, 4 January 2023
- The following problem is from both the 2022 AMC 10B #6 and 2022 AMC 12B #3, so both problems redirect to this page.
Contents
Problem
How many of the first ten numbers of the sequence 
are prime numbers?
Solution 1 (Generalization)
The 
th term of this sequence is
![\[\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.\]](http://latex.artofproblemsolving.com/d/5/b/d5b8d71075dff9c87a43c0c4dd2d32c4c3b1415d.png)
It follows that the terms are
Therefore, there are 
prime numbers in this sequence.
~MRENTHUSIASM
Solution 2 (Simple Sums)
Observe how
all take the form of ![\[\overbrace{111\ldots}^{n+1}\overbrace{00\ldots}^{n} + \overbrace{111\ldots}^{n+1} = \overbrace{111\ldots}^{n+1}(10^{n} + 1).\]](http://latex.artofproblemsolving.com/1/8/0/180a93702c02bc7be9e6821fbc7425efbfdf49bc.png)
Factoring each of the sums, we have ![\[11(10+1), 111(100+1), 1111(1000+1), \ldots\]](http://latex.artofproblemsolving.com/6/4/f/64f92bf0e88c9fd5624d4a464589723aab6a048b.png)
respectively. With each number factored, there are primes in the set.
~ab2024
Solution 3 (Educated Guesses)
Note that it's obvious that 
is divisible by 
and 
is divisible by 
therefore, since this an AMC 10 problem 6, we may safely assume that we do not need to check two-digit prime divisibility or use obscure theorems. So, the answer is 
~Dhillonr25
See Also
| 2022 AMC 10B (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 10 Problems and Solutions | ||
| 2022 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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. 
