Difference between revisions of "2022 AMC 10B Problems/Problem 6"
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~ab2024 | ~ab2024 | ||
| − | ==Solution 3 (Educated | + | ==Solution 3 (Educated Guessing)== |
| − | Note | + | Note that <math>121</math> is divisible by <math>11</math> and <math>11211</math> is divisible by <math>3</math>. Hence, because this problem 6 of the AMC 10, we assume we do not need to check two-digit prime divisibility or use obscure theorems. Therefore, the answer is <math>\boxed{\textbf{(A) } 0}.</math> |
~Dhillonr25 | ~Dhillonr25 | ||
Revision as of 01:01, 29 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 Guessing)
Note that 
is divisible by 
and 
is divisible by 
. Hence, because this problem 6 of the AMC 10, we assume we do not need to check two-digit prime divisibility or use obscure theorems. Therefore, 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. 
