Library / Slovenia — National Math Olympiad /5 of 122

Slovenia — National Math Olympiad

Algebra Difficulty 3.8 AMC 10/12 Find the answer Slovenia

Let $m$ and $n$ be positive integers, such that $19 \le m \le 49$ , $51 \le n \le 101$ . What is the greatest possible value of the expression $\frac{n+m}{n-m}$ ?

Pick one

Solution

We have $\frac{n+m}{n-m} = \frac{n-m+2m}{n-m} = 1 + 2\frac{m}{n-m}$ , so the value is maximal when $n-m = 2$ and $m = 49$ . Then, the value is $50$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.