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

(New page: == 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...) |
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== 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>. |

Now in order for <math>x^{\lfloor x\rfloor}</math> to be an integer, <math>x</math> must be an integral root of an integer, | Now in order for <math>x^{\lfloor x\rfloor}</math> to be an integer, <math>x</math> must be an integral root of an integer, | ||

+ | |||

+ | So let do case work: | ||

+ | |||

+ | For <math>{\lfloor x\rfloor}=0</math> N=<math>1</math> no matter what x is | ||

+ | |||

+ | For <math>{\lfloor x\rfloor}=1</math> N can be anything between <math>1^1</math> to <math>2^1</math> excluding <math>2^1</math> | ||

+ | |||

+ | This gives us <math>2^1-1^1=1</math> N's | ||

+ | |||

+ | For <math>{\lfloor x\rfloor}=2</math> N can be anything between <math>2^2</math> to <math>3^2</math> excluding <math>3^2</math> | ||

+ | |||

+ | This gives us <math>3^2-2^2=5</math> N's | ||

+ | |||

+ | For <math>{\lfloor x\rfloor}=3</math> N can be anything between <math>3^3</math> to <math>4^3</math> excluding <math>4^3</math> | ||

+ | |||

+ | This gives us <math>4^3-3^3=37</math> N's | ||

+ | |||

+ | For <math>{\lfloor x\rfloor}=4</math> N can be anything between <math>4^4</math> to <math>5^4</math> excluding <math>5^4</math> | ||

+ | |||

+ | This gives us <math>5^4-4^4=369</math> N's | ||

+ | |||

+ | Since <math>x</math> must be less than <math>5</math>, we can stop here | ||

+ | |||

+ | Answer <math>= 1+5+37+369= \boxed {412}</math> |

## Revision as of 22:43, 19 March 2009

## 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 .

Now in order for to be an integer, must be an integral root of an integer,

So let do case work:

For N= no matter what x is

For N can be anything between to excluding

This gives us N's

For N can be anything between to excluding

This gives us N's

For N can be anything between to excluding

This gives us N's

For N can be anything between to excluding

This gives us N's

Since must be less than , we can stop here

Answer