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Let $n$ is a natural number. Prove that $$\left\lfloor\frac{n}{1}\right\rfloor+\left\lfloor\frac{n}{2}\right\rfloor+....+\left\lfloor\frac{n}{n}\right\rfloor+\left\lfloor{\sqrt{n}}\right\rfloor$$
is even.

I used induction and the inequality $$x-1<\lfloor{x}\rfloor\le{x}$$ to prove it.

Is there any other, nicer way to do it?

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Induction is a good way to go and the inequality you used is all we have about the floor function, so what could be a better way? – Hagen von Eitzen Feb 2 '14 at 18:27
1  
A duplicate of this question was asked later. You might find my answer to that question useful. – David Feb 18 '15 at 4:58
    
@David +1. Truly remarkable answer by you in that post. – Ishan Singh Feb 18 '15 at 7:36
up vote 10 down vote accepted

Let $$S = \{\,(a,b)\in\mathbb N^2\mid ab\le n\,\} $$ and $$ T=\{\,(a,a)\in\mathbb N^2\mid a^2\le n\,\}.$$ Then $$ |S|=\sum_{b\ge 1}|\{\,a\in\mathbb N\mid a\le\tfrac nb\,\}|=\sum_{b\ge 1}\lfloor \tfrac nb\rfloor =\left\lfloor \frac n1\right\rfloor+\left\lfloor \frac n2\right\rfloor+\ldots +\left\lfloor \frac nn\right \rfloor$$ and $$ |T|=\lfloor \sqrt n\rfloor.$$ The map $(a,b)\mapsto (b.a)$ is a fixedpoint-free involution of $S\setminus T$, hence $|S\setminus T|$ is even. Since $$ \left\lfloor \frac n1\right\rfloor+\left\lfloor \frac n2\right\rfloor+\ldots +\left\lfloor \frac nn\right \rfloor+\lfloor\sqrt n\rfloor = |S\setminus T|+2|T|$$ we are done.

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