# 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 $x$? (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 ${\lfloor x\rfloor}=0$, $N=1$ 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.

## Video Solution

Mostly the above solution explained on video: https://www.youtube.com/watch?v=2Xzjh6ae0MU&t=11s

~IceMatrix

~Shreyas S