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Find out the number of integer solution of $[\frac{x}{100}[\frac{x}{100}]]=5$
My Attempt
let $x=100k+t$ where $0 \leq t \leq 99$ and $k>=0$. Thus $[\frac{x}{100}]=k$. Hence $[\frac{x}{100}k]=5$ thus $\frac{x}{100}k=5+ r$ where $0 \leq r < 1$
From this point I am stuck what to do. Can anyone help? Thanks in advance.

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    $\begingroup$ Isn't this non-decreasing? Find the lowest $x$ and the greatest $x$ that satisfy the condition, and everything else in between will work too, won't it? $\endgroup$ – Brian Tung Oct 18 '16 at 5:33
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Try it with $y=x/100$ first. [y [y]]=5. We know $ [y][y]\le [y [y]]=5\le y [y]<([y]+1)[y] $. From that we can see $[y]^2 \le 5 <[y]^2+[y] $.

From this we can conclude $[y] =2$.

And from there it solves itself.

$[y [y]]=[2y]=5$

$5 \le 2y < 5+1$

$5/2\le y < 3$

$5/2 \le x/100 <3$

$250\le x < 300$

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