# 2009 AIME I Problems/Problem 6

## 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 $1^1,\ 2^2,\ 3^3,\ 4^4,\ \sqrt{5}^{\lfloor\sqrt{5}\rfloor},\ \sqrt{6}^{\lfloor\sqrt{6}\rfloor},\ \sqrt{7}^{\lfloor\sqrt{7}\rfloor},\ \sqrt{8}^{\lfloor\sqrt{8}\rfloor},\ \sqrt{28}^{\lfloor\sqrt{28}\rfloor},\ \sqrt{29}^{\lfloor\sqrt{29}\rfloor},\ \sqrt{30}^{\lfloor\sqrt{30}\rfloor},\ ...,$ $\ \sqrt{63}^{\lfloor\sqrt{63}\rfloor},\ \sqrt{257}^{\lfloor\sqrt{257}\rfloor},\ \sqrt{258}^{\lfloor\sqrt{258}\rfloor},\ ...,\ \sqrt{624}^{\lfloor\sqrt{624}\rfloor}$ 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

~Shreyas S

### Video Solution 3

Projective Solution: https://youtu.be/fUef_tVnM5M

~Shreyas S

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 