Help, RMO sample question

Define $\displaystyle q(n) = \lfloor \frac{n}{\lfloor \sqrt{n} \rfloor}\rfloor$ for $n=1,2,3,\ldots$ . Find all $n$ satisfying $q(n) > q(n+1)$ .

Notation: $\lfloor \cdot \rfloor$ denotes the floor function.

#Algebra

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

The solutions are the natural numbers one less than a perfect square, i.e., the solution set is $S=\{k^2-1\mid k\in\Bbb N\setminus\{1\}\}=\{3,8,15,\ldots\}$

To prove this, you need to prove that $q(n)\gt q(n+1)$ when $n\in S$ and $q(n)\leq q(n+1)$ otherwise.

For the first part, note that when $n=k^2-1$ , since the square root function is strictly increasing, we have $\lfloor\sqrt n\rfloor=k-1$ which implies that $q(n)=k+1$ whereas we have $\lfloor\sqrt{n+1}\rfloor=k$ which implies that $q(n+1)=k$ which shows that $q(n)=k+1\gt k=q(n+1)$ when $n\in S$ .

Now, for the second part, when $n\notin S$ , we note that when $k^2\leq n\lt (k+1)^2-1$ , we have $\lfloor\sqrt n\rfloor=\lfloor\sqrt{n+1}\rfloor=k$ which implies that $q(n)=\lfloor\frac nk\rfloor$ and $q(n+1)=\lfloor\frac{n+1}k\rfloor$ . Now, since $\frac nk\lt\frac {n+1}k$ , applying the floor function on both sides of the inequality, we have $\lfloor\frac nk\rfloor\leq\lfloor\frac{n+1}k\rfloor$ from which we conclude that $q(n)\leq q(n+1)$ .

Prasun Biswas - 4 years, 8 months ago

Brilliant Solution!+1

Ayush G Rai - 4 years, 8 months ago

abhishek alva - 4 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$