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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$.)

It will either be:

$x^1, x^2, x^3, x^4, ..., x^k$ for an integer $k$.

$$x^k - N = 0$$

I mean I am sort of confused because:

$$x = \sqrt[k]{N}$$

But the requirement is that:

$$k \le \sqrt[k]{N} < k + 1$$

For $k = 2, N = 4,$ it follows, $2 = 2 < 3$. That is one solution.

$$k^k \le N < (k + 1)^k$$

For $N$ there are: $(k + 1)^k - k^k$ solutions.

But there are infinite $k's$ so I am confused.

HINTS ONLY PLEASE!!

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Hint: There are not infinite k's. First note that evidently $\lfloor x \rfloor$ is an integer. Also, for a certain integer $\lfloor x \rfloor$, $x^{\lfloor x \rfloor}$ exceeds $1000$, and is hence not viable. You must only consider the integer cases for $\lfloor x \rfloor$ which are smaller than this value. Once you know this limiting $\lfloor x \rfloor$, it is relatively easy to count the number of viable solutions for each value of $\lfloor x \rfloor $ smaller than it by considering possible values of $N$ which fall within some given interval.

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  • $\begingroup$ I think I understand what @DonkeyKong meant. Say for integer $\lfloor x\rfloor = 2$, list all possible integer values from $2^2$ up to, but not including, $(2+1)^2$. Then go on the next one $\lfloor x\rfloor = 3$. $\endgroup$ – peterwhy Jul 15 '15 at 21:28
  • $\begingroup$ Yeah I see, $k=5$ does the trick (+1) $\endgroup$ – Amad27 Jul 16 '15 at 10:53
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Using the definition of the floor function:

$N \in [1^1,\lfloor1.999^1\rfloor] \cup [2^2,\lfloor2.999^2\rfloor] \cup [3^3,\lfloor3.999^3\rfloor] \cup [4^4,\lfloor4.999^4\rfloor]$

[edit] or better yet:

$N \in [1,1] \cup [2^2,3^2-1] \cup [3^3,4^3-1] \cup [4^4,5^4-1]$

And $5^5 > 1000$

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