Difference between revisions of "2022 AMC 10B Problems/Problem 6"
MRENTHUSIASM (talk | contribs) (Created page with "==Problem== How many of the first ten numbers of the sequence <math>121, 11211, 1112111, \ldots</math> are prime numbers? <math>\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \...") |
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==Solution== | ==Solution== | ||
− | The | + | The <math>n</math>th term of this sequence is |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | + | \left(10^{2n}+10^{2n-1}+\cdots+10^{n}\right)+\left(10^{n}+10^{n-1}+\cdots+10^{0}\right) &= 10^n\left(10^{n}+10^{n-1}+\cdots+10^{0}\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 first ten terms are divisible by <math>11,101,1001,\ldots,</math> respectively. | ||
+ | |||
+ | Therefore, there are \boxed{\textbf{(A) } 0} prime numbers in this sequence. | ||
+ | |||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 16:30, 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 first ten terms are divisible by
respectively.
Therefore, there are \boxed{\textbf{(A) } 0} 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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.