Difference between revisions of "2022 AMC 10B Problems/Problem 6"
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&= \left(10^n + 1\right)\left(10^{n}+10^{n-1}+\cdots+10^{0}\right). | &= \left(10^n + 1\right)\left(10^{n}+10^{n-1}+\cdots+10^{0}\right). | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | It follows that the | + | It follows that the terms are |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
121 &= 11\cdot11, \ | 121 &= 11\cdot11, \ | ||
11211 &= 101\cdot111, \ | 11211 &= 101\cdot111, \ | ||
| − | 1112111 &= 1001\cdot1111 | + | 1112111 &= 1001\cdot1111, \ |
| + | & \ \vdots | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
Therefore, there are <math>\boxed{\textbf{(A) } 0}</math> prime numbers in this sequence. | Therefore, there are <math>\boxed{\textbf{(A) } 0}</math> prime numbers in this sequence. | ||
Revision as of 16:35, 17 November 2022
Problem
How many of the first ten numbers of the sequence 
are prime numbers?
Solution
The 
th term of this sequence is
It follows that the terms are
Therefore, there are 
prime numbers in this sequence.
~MRENTHUSIASM
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
