Difference between revisions of "2014 AIME II Problems/Problem 15"
m (always that i before e except after c rule...) |
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<math>x_7 = 30</math> <math>x_{15} = 210</math> | <math>x_7 = 30</math> <math>x_{15} = 210</math> | ||
− | Commit to the bash. Eventually, you will | + | Commit to the bash. Eventually, you will receive that <math>x_{149} = 2090</math>, so <math>\boxed{149}</math> is the answer. Trust me, this is worth the 10 index points. |
<math>\textbf{-RootThreeOverTwo}</math> | <math>\textbf{-RootThreeOverTwo}</math> | ||
+ | |||
== See also == | == See also == | ||
{{AIME box|year=2014|n=II|num-b=14|after=Last Problem}} | {{AIME box|year=2014|n=II|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:44, 25 March 2020
Problem
For any integer , let be the smallest prime which does not divide Define the integer function to be the product of all primes less than if , and if Let be the sequence defined by , and for Find the smallest positive integer such that
Solution
Note that for any , for any prime , . This provides motivation to translate into a binary sequence .
Let the prime factorization of be written as , where is the th prime number. Then, for every in the prime factorization of , place a in the th digit of . This will result in the conversion .
Multiplication for the sequence will translate to addition for the sequence . Thus, we see that translates into . Since , and , corresponds to , which is in binary. Since , = .
Solution 2 (Painful Bash)
We go through the terms and look for a pattern. We find that
Commit to the bash. Eventually, you will receive that , so is the answer. Trust me, this is worth the 10 index points.
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.