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Find $$\sum_{n=1}^{2016}\left\lfloor\frac{\sqrt n}{10}\right\rfloor$$ where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.

Apparently it is to be solved within 2 minutes! But I am completely lost on how to solve this.

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closed as off-topic by user21820, GNUSupporter 8964民主女神 地下教會, Did, Claude Leibovici, Xander Henderson Mar 12 '18 at 13:15

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    $\begingroup$ Hint: $0+0+0+0+\cdots+0+1+1+1+1+\cdots+1+\cdots+2+2+2+2+\cdots$ $\endgroup$ – Yves Daoust Sep 7 '16 at 8:17
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Just write it as:

$$\sum \left\lfloor{\frac{\sqrt{n}}{10}}\right\rfloor=\sum \left\lfloor \sqrt{\frac{n}{100}}\right\rfloor$$

Then just search for when you change integer, for example you get $3$ as soon as you are above $\sqrt{9}$ and under $\sqrt{16}$ where you will change for $4$, so for $n\geq 900$, $n<1600$.

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  • $\begingroup$ Pretty smart idea. +1 $\endgroup$ – DonAntonio Sep 7 '16 at 8:25
  • $\begingroup$ Very clever! (+1) $\endgroup$ – hypergeometric Sep 7 '16 at 14:58
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Look at the places where $f(n)=\frac{\sqrt n}{10}$ "trips" across integers:

  • $f(n)=1$ when $n=100$, so $\lfloor f(n)\rfloor=0$ when $n<100$
  • $f(n)=2$ when $n=400$, so $\lfloor f(n)\rfloor=1$ when $100\le n<400$ (300 numbers)
  • Similarly, $\lfloor f(n)\rfloor=2$ when $400\le n<900$ (500 numbers)…
  • $\lfloor f(n)\rfloor=3$ when $900\le n<1600$ (700 numbers)…
  • $\lfloor f(n)\rfloor=4$ when $1600\le n<2017$ (417 numbers).

The answer is $$300+2×500+3×700+4×417=5068$$

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    $\begingroup$ I got the same answer by the same approach. It took me 1 minute 53 seconds. $\endgroup$ – Barry Cipra Sep 7 '16 at 11:36
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Notice that $$ \frac{\sqrt{n}}{10} \le \frac{\sqrt{2016}}{10} < 5. $$ That means, that for $1 \le n \le 2016$, we have $$ \left\lfloor\frac{\sqrt{n}}{10} \right\rfloor \in \{0, 1, 2, 3, 4\}. $$ You just have to count for how many $n$ it is each of $0, 1, 2, 3$, and $4$.

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