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Difference between revisions of "2009 AIME I Problems/Problem 6"

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== Problem ==
 
== Problem ==
  
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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>? (The notation <math>\lfloor x\rfloor</math> denotes the greatest integer that is less than or equal to <math>x</math>.)
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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>?
  
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== Solution ==
+
== Solution 1==
 
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>.  
  
−
Now, <math>{\lfloor x\rfloor}</math> must be an integer, so lets do case work:
+
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> no matter what <math>x</math> is
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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}=1</math>, <math>N</math> can be anything between <math>1^1</math> to <math>2^1</math> excluding <math>2^1</math>
 
For <math>{\lfloor x\rfloor}=1</math>, <math>N</math> can be anything between <math>1^1</math> to <math>2^1</math> excluding <math>2^1</math>
  
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This gives us <math>2^1-1^1=1</math> <math>N</math>'s
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Therefore, <math>N=1</math>. However, we got <math>N=1</math> in case 1 so it got counted twice.
  
 
For <math>{\lfloor x\rfloor}=2</math>, <math>N</math> can be anything between <math>2^2</math> to <math>3^2</math> excluding <math>3^2</math>
 
For <math>{\lfloor x\rfloor}=2</math>, <math>N</math> can be anything between <math>2^2</math> to <math>3^2</math> excluding <math>3^2</math>
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This gives us <math>5^4-4^4=369</math> <math>N</math>'s
 
This gives us <math>5^4-4^4=369</math> <math>N</math>'s
  
−
Since <math>x</math> must be less than <math>5</math>, we can stop here and the answer answer is <math>\boxed {4}</math> possible values for <math>N</math>.
+
Since <math>x</math> must be less than <math>5</math>, we can stop here and the answer is <math>1+5+37+369= \boxed {412}</math> possible values for <math>N</math>.
 +
 
 +
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},\ ...,</math>
 +
<math>\ \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 15:38, 15 February 2021

Problem

How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$?

Solution 1

First, $x$ must be less than $5$, since otherwise $x^{\lfloor x\rfloor}$ would be at least $3125$ which is greater than $1000$.

Because ${\lfloor x\rfloor}$ must be an integer, let’s do case work based on ${\lfloor x\rfloor}$:

For ${\lfloor x\rfloor}=0$, $N=1$ as long as $x \neq 0$. This gives us $1$ value of $N$.

For ${\lfloor x\rfloor}=1$, $N$ can be anything between $1^1$ to $2^1$ excluding $2^1$

Therefore, $N=1$. However, we got $N=1$ in case 1 so it got counted twice.

For ${\lfloor x\rfloor}=2$, $N$ can be anything between $2^2$ to $3^2$ excluding $3^2$

This gives us $3^2-2^2=5$ $N$'s

For ${\lfloor x\rfloor}=3$, $N$ can be anything between $3^3$ to $4^3$ excluding $4^3$

This gives us $4^3-3^3=37$ $N$'s

For ${\lfloor x\rfloor}=4$, $N$ can be anything between $4^4$ to $5^4$ excluding $5^4$

This gives us $5^4-4^4=369$ $N$'s

Since $x$ must be less than $5$, we can stop here and the answer is $1+5+37+369= \boxed {412}$ possible values for $N$.

Alternatively, one could find that the values which work are $1^1,\ 2^2,\ 3^3,\ 4^4,\ \sqrt{5}^{\lfloor\sqrt{5}\rfloor},\ \sqrt{6}^{\lfloor\sqrt{6}\rfloor},\ \sqrt{7}^{\lfloor\sqrt{7}\rfloor},\ \sqrt{8}^{\lfloor\sqrt{8}\rfloor},\ \sqrt[3]{28}^{\lfloor\sqrt[3]{28}\rfloor},\ \sqrt[3]{29}^{\lfloor\sqrt[3]{29}\rfloor},\ \sqrt[3]{30}^{\lfloor\sqrt[3]{30}\rfloor},\ ...,$ $\ \sqrt[3]{63}^{\lfloor\sqrt[3]{63}\rfloor},\ \sqrt[4]{257}^{\lfloor\sqrt[4]{257}\rfloor},\ \sqrt[4]{258}^{\lfloor\sqrt[4]{258}\rfloor},\ ...,\ \sqrt[4]{624}^{\lfloor\sqrt[4]{624}\rfloor}$ to get the same answer.

Solution 2

For a positive integer $k$, we find the number of positive integers $N$ such that $x^{\lfloor x\rfloor}=N$ has a solution with ${\lfloor x\rfloor}=k$. Then $x=\sqrt[k]{N}$, and because $k \le x < k+1$, we have $k^k \le x^k < (k+1)^k$, and because $(k+1)^k$ is an integer, we get $k^k \le x^k \le (k+1)^k-1$. The number of possible values of $x^k$ is equal to the number of integers between $k^k$ and $(k+1)^k-1$ inclusive, which is equal to the larger number minus the smaller number plus one or $((k+1)^k-1)-(k^k)+1$, and this is equal to $(k+1)^k-k^k$. If $k>4$, the value of $x^k$ exceeds $1000$, so we only need to consider $k \le 4$. The requested number of values of $N$ is the same as the number of values of $x^k$, which is $\sum^{4}_{k=1} [(k+1)^k-k^k]=2-1+9-4+64-27+625-256=\boxed{412}$.

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

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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