Difference between revisions of "2009 AIME I Problems/Problem 6"
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== Problem == | == Problem == | ||
− | How many positive integers <math>N</math> less than <math>1000</math> are there such that the equation <math>x^{\lfloor x\rfloor} = N</math> has a solution for <math>x</math>? | + | How many positive integers <math>N</math> less than <math>1000</math> are there such that the equation <math>x^{\lfloor x\rfloor} = N</math> has a solution for <math>x</math>? |
== Solution == | == Solution == | ||
First, <math>x</math> must be less than <math>5</math>, since otherwise <math>x^{\lfloor x\rfloor}</math> would be at least <math>3125</math> which is greater than <math>1000</math>. | First, <math>x</math> must be less than <math>5</math>, since otherwise <math>x^{\lfloor x\rfloor}</math> would be at least <math>3125</math> which is greater than <math>1000</math>. | ||
− | Because <math>{\lfloor x\rfloor}</math> must be an integer, | + | Because <math>{\lfloor x\rfloor}</math> must be an integer, let’s do case work based on <math>{\lfloor x\rfloor}</math>: |
For <math>{\lfloor x\rfloor}=0</math>, <math>N=1</math> as long as <math>x \neq 0</math>. This gives us <math>1</math> value of <math>N</math>. | For <math>{\lfloor x\rfloor}=0</math>, <math>N=1</math> as long as <math>x \neq 0</math>. This gives us <math>1</math> value of <math>N</math>. | ||
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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. | ||
+ | |||
+ | ==Solution 2== | ||
+ | For a positive integer <math>k</math>, we find the number of positive integers <math>N</math> such that <math>x^{\lfloor x\rfloor}=N</math> has a solution with <math>{\lfloor x\rfloor}=k</math>. Then <math>x=\sqrt[k]{N}</math>, and because <math>k \le x < k+1</math>, we have <math>k^k \le x^k < (k+1)^k</math>, and because <math>(k+1)^k</math> is an integer, we get <math>k^k \le x^k \le (k+1)^k-1</math>. The number of possible values of <math>x^k</math> is equal to the number of integers between <math>k^k</math> and <math>(k+1)^k-1</math> inclusive, which is equal to the larger number minus the smaller number plus one or <math>((k+1)^k-1)-(k^k)+1</math>, and this is equal to <math>(k+1)^k-k^k</math>. If <math>k>4</math>, the value of <math>x^k</math> exceeds <math>1000</math>, so we only need to consider <math>k \le 4</math>. The requested number of values of <math>N</math> is the same as the number of values of <math>x^k</math>, which is <math> \sum^{4}_{k=1} [(k+1)^k-k^k]=2-1+9-4+64-27+625-256=\boxed{412}</math>. | ||
+ | |||
+ | ==Video Solutions== | ||
+ | |||
+ | ===Video Solution 1=== | ||
+ | Mostly the above solution explained on video: https://www.youtube.com/watch?v=2Xzjh6ae0MU&t=11s | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ===Video Solution 2=== | ||
+ | https://youtu.be/kALrIDMR0dg | ||
+ | |||
+ | ~Shreyas S | ||
+ | |||
+ | ===Video Solution 3=== | ||
+ | Projective Solution: https://youtu.be/fUef_tVnM5M | ||
+ | |||
+ | ~Shreyas S | ||
== 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}} | ||
+ | [[Category:Intermediate Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:18, 9 October 2020
Contents
Problem
How many positive integers less than are there such that the equation has a solution for ?
Solution
First, must be less than , since otherwise would be at least which is greater than .
Because must be an integer, let’s do case work based on :
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.
Solution 2
For a positive integer , we find the number of positive integers such that has a solution with . Then , and because , we have , and because is an integer, we get . The number of possible values of is equal to the number of integers between and inclusive, which is equal to the larger number minus the smaller number plus one or , and this is equal to . If , the value of exceeds , so we only need to consider . The requested number of values of is the same as the number of values of , which is .
Video Solutions
Video Solution 1
Mostly the above solution explained on video: https://www.youtube.com/watch?v=2Xzjh6ae0MU&t=11s
~IceMatrix
Video Solution 2
~Shreyas S
Video Solution 3
Projective Solution: https://youtu.be/fUef_tVnM5M
~Shreyas S
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 |
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