Difference between revisions of "2009 AIME I Problems/Problem 6"
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Alternatively, one could find that the values which work are <math>1^1,\ 2^2,\ 3^3,\ 4^4,\ \sqrt{5}^{\lfloor\sqrt{5}\rfloor},\ \sqrt{6}^{\lfloor\sqrt{6}\rfloor},\ \sqrt{7}^{\lfloor\sqrt{7}\rfloor},\ \sqrt{8}^{\lfloor\sqrt{8}\rfloor},\ \sqrt[3]{28}^{\lfloor\sqrt[3]{28}\rfloor},\ \sqrt[3]{29}^{\lfloor\sqrt[3]{29}\rfloor},\ \sqrt[3]{30}^{\lfloor\sqrt[3]{30}\rfloor},\ ...,\ \sqrt[3]{63}^{\lfloor\sqrt[3]{63}\rfloor},\ \sqrt[4]{257}^{\lfloor\sqrt[4]{257}\rfloor},\ \sqrt[4]{258}^{\lfloor\sqrt[4]{258}\rfloor},\ ...,\ \sqrt[4]{624}^{\lfloor\sqrt[4]{624}\rfloor}</math> to get the same answer. | Alternatively, one could find that the values which work are <math>1^1,\ 2^2,\ 3^3,\ 4^4,\ \sqrt{5}^{\lfloor\sqrt{5}\rfloor},\ \sqrt{6}^{\lfloor\sqrt{6}\rfloor},\ \sqrt{7}^{\lfloor\sqrt{7}\rfloor},\ \sqrt{8}^{\lfloor\sqrt{8}\rfloor},\ \sqrt[3]{28}^{\lfloor\sqrt[3]{28}\rfloor},\ \sqrt[3]{29}^{\lfloor\sqrt[3]{29}\rfloor},\ \sqrt[3]{30}^{\lfloor\sqrt[3]{30}\rfloor},\ ...,\ \sqrt[3]{63}^{\lfloor\sqrt[3]{63}\rfloor},\ \sqrt[4]{257}^{\lfloor\sqrt[4]{257}\rfloor},\ \sqrt[4]{258}^{\lfloor\sqrt[4]{258}\rfloor},\ ...,\ \sqrt[4]{624}^{\lfloor\sqrt[4]{624}\rfloor}</math> to get the same answer. | ||
+ | |||
+ | ==Video Solution== | ||
+ | Mostly the above solution explained on video: https://www.youtube.com/watch?v=2Xzjh6ae0MU&t=11s | ||
+ | |||
+ | ~IceMatrix | ||
== See also == | == See also == | ||
{{AIME box|year=2009|n=I|num-b=5|num-a=7}} | {{AIME box|year=2009|n=I|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:05, 21 February 2020
Contents
[hide]Problem
How many positive integers less than are there such that the equation has a solution for ? (The notation denotes the greatest integer that is less than or equal to .)
Solution
First, must be less than , since otherwise would be at least which is greater than .
Because must be an integer, we can do some simple case work:
For , as long as . This gives us value of .
For , can be anything between to excluding
Therefore, . However, we got in case 1 so it got counted twice.
For , can be anything between to excluding
This gives us 's
For , can be anything between to excluding
This gives us 's
For , can be anything between to excluding
This gives us 's
Since must be less than , we can stop here and the answer is possible values for .
Alternatively, one could find that the values which work are to get the same answer.
Video Solution
Mostly the above solution explained on video: https://www.youtube.com/watch?v=2Xzjh6ae0MU&t=11s
~IceMatrix
See also
2009 AIME I (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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.