# $\sum_{n=1}^{2016}\left\lfloor\frac{\sqrt n}{10}\right\rfloor$ in 2 minutes [closed]

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.

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

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• Hint: $0+0+0+0+\cdots+0+1+1+1+1+\cdots+1+\cdots+2+2+2+2+\cdots$ – Yves Daoust Sep 7 '16 at 8:17

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 $n\geq 900$, $n<1600$.

• Pretty smart idea. +1 – DonAntonio Sep 7 '16 at 8:25
• Very clever! (+1) – hypergeometric Sep 7 '16 at 14:58

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 $$300+2×500+3×700+4×417=5068$$

• I got the same answer by the same approach. It took me 1 minute 53 seconds. – Barry Cipra Sep 7 '16 at 11:36

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 $4$.