Difference between revisions of "2022 AMC 12B Problems/Problem 3"
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Observe that <cmath>\overline{P(1,n) 2 P(1,n)} = \overline{P(1,n+1)P(0,n)} + P(1,n+1) = P(1,n+1) \cdot 10^n + P(1,n+1) = (P(1,n+1))(10^n + 1).</cmath> | Observe that <cmath>\overline{P(1,n) 2 P(1,n)} = \overline{P(1,n+1)P(0,n)} + P(1,n+1) = P(1,n+1) \cdot 10^n + P(1,n+1) = (P(1,n+1))(10^n + 1).</cmath> | ||
Clearly, both terms are larger than <math>1</math> since <math>n \geq 1</math>, hence all the numbers of the sequence are <math>\fbox{0(A)}</math>, and we're done! | Clearly, both terms are larger than <math>1</math> since <math>n \geq 1</math>, hence all the numbers of the sequence are <math>\fbox{0(A)}</math>, and we're done! | ||
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~[[User:Bxiao31415|Bxiao31415]] | ~[[User:Bxiao31415|Bxiao31415]] | ||
Revision as of 17:18, 17 November 2022
Problem
How many of the first ten numbers of the sequence , , , ... are prime numbers?
Solution 1
Let denote the digit written times and let denote the concatenation of , , ..., . Observe that Clearly, both terms are larger than since , hence all the numbers of the sequence are , and we're done!
See Also
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.