No one?

Is $[a]$ the floor function or the nearest integer?

[x] means the greateat integer which is smaller than x or equal to x.Maybe my English is poor.

I know a similar problem. I have solved it by induction, therefore we are ready if we had shown that $x[\frac{k}{x}]-(x+1)[\frac{k}{x+1}]\leq 1$ I try to prove (or disprove) this inequality.

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Gauß wrote: I know a similar problem. I have solved it by induction, therefore we are ready if we had shown that $x[\frac{k}{x}]-(x+1)[\frac{k}{x+1}]\leq 1$ I try to prove (or disprove) this inequality. It is not right.put x=k

Yes you're right.

Is this so hard?

can we prove $x[\frac{k-1}{x}]-(x+1)[\frac{k}{x+1}]\leq 0$?

I have thought about it and I want to give my results. (i) $k-x\leq x\left[ \frac{k}{x} \right] \leq k$ My conjecture is that we have to regard the expression: $\sum_{k=1}^n \left[ \frac{k}{x} \right] $ Because of the monotony we must detremine the intervall $[a,b] $ for wich $\left[ \frac{k}{x} \right]$ is constant. $qx\leq k \leq (q+1)x \Rightarrow \left[ \frac{k}{x} \right]$ is for at most x values constant. These are my results, so far. I'm lookink for more results yet. These are only notes, wich I want to post.

It's seems the problem's text disappeared Can anyone formulate it again?

The problem statement can be found in the pdf file in the second post here: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=681379332&t=41023.

It seems that it's still unsolved!

You can see the official solutions Here But I'm not sure if it's understandable.