Maths Olympiad Prep

Library / /5 of 40

Algebra Difficulty 5.3 AIME, harder Prove it China

Find the minimum positive integer nn such that
n−20112012−n−20122011<n−201320113−n−201120133 \sqrt{\frac{n-2011}{2012}} - \sqrt{\frac{n-2012}{2011}} < \sqrt[3]{\frac{n-2013}{2011}} - \sqrt[3]{\frac{n-2011}{2013}}

Solution

We see that if 2012≤n≤40232012 \le n \le 4023, then n−20112012−n−20122011≥0\sqrt{\frac{n-2011}{2012}} - \sqrt{\frac{n-2012}{2011}} \ge 0 and n−201320113−n−201120133<0\sqrt[3]{\frac{n-2013}{2011}} - \sqrt[3]{\frac{n-2011}{2013}} < 0.

Otherwise,
n−20112012≤n−20122011⇔n>4023n−201320113≥n−201120133⇔n≥4024. \begin{aligned} \sqrt{\frac{n-2011}{2012}} &\le \sqrt{\frac{n-2012}{2011}} &&\Leftrightarrow n > 4023 \\ \sqrt[3]{\frac{n-2013}{2011}} &\ge \sqrt[3]{\frac{n-2011}{2013}} &&\Leftrightarrow n \ge 4024. \end{aligned}
Thus, if n≥4024n \ge 4024, then
n−20112012−n−20122011<0≤n−201320113−n−201120133 \sqrt{\frac{n-2011}{2012}} - \sqrt{\frac{n-2012}{2011}} < 0 \le \sqrt[3]{\frac{n-2013}{2011}} - \sqrt[3]{\frac{n-2011}{2013}}
So the minimum of nn is 40244024. 4024\boxed{4024}

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.