Difference between revisions of "2014 AIME II Problems/Problem 15"
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==Solution== | ==Solution== | ||
| − | Note that (in base 2 for the indices, base 10 for the values) <math>a_1 = 2, a_{10} = 3, a_{11} = 2 * 3 = 6, ..., a_{10010101} = 2 | + | Note that (in base 2 for the indices, base 10 for the values) <math>a_1 = 2, a_{10} = 3, a_{11} = 2 * 3 = 6, ..., a_{10010101} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090.</math> Thus, <math>t = 10010101</math> (base 2) = <math>\boxed{149}</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 21:46, 20 April 2015
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 (in base 2 for the indices, base 10 for the values) 
Thus, 
(base 2) = 
.
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. 
