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

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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
  
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Since <math>x</math> must be less than <math>5</math>, we can stop here and the answer is <math>\boxed {4}</math> possible values for <math>N</math>.
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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>.
  
 
== 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 22:43, 27 January 2015

Problem

How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$? (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$.)

Solution

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, we can do some simple case work:

For ${\lfloor x\rfloor}=0$, $N=1$ no matter what $x$ is

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

This gives us $2^1-1^1=1$ $N$'s

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 $N$.

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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