2009 AIME I Problems/Problem 6

Revision as of 14:05, 20 March 2009 by Kubluck (talk | contribs) (Solution)

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 $1000$.

Now in order for $x^{\lfloor x\rfloor}$ to be an integer, $x$ must be an integral root of an integer, so lets do 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 answer is $1+5+37+369= \boxed {412}$.

See also

2009 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AIME Problems and Solutions
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