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! | ||
| + | ~[[User:Bxiao31415|Bxiao31415]] | ||
| + | |||
| + | == See Also == | ||
| + | {{AMC12 box|year=2022|ab=B|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
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 ![\[\overline{P(1,n) 2 P(1,n)} = \overline{P(1,n+1)P(0,n)} + P(1,n+1) = P(1,n+1) \cdot 10^n + P(1,n+1) = (P(1,n+1))(10^n + 1).\]](http://latex.artofproblemsolving.com/a/9/8/a98263e7ccde384634ef4b630aa3715a5665cf56.png)
Clearly, both terms are larger than 
since 
, hence all the numbers of the sequence are 
, and we're done!
~Bxiao31415
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. 
