Competition problem — Number theory

Number theory Difficulty 4.5 AIME Prove it Brazil

Show that for any positive integer $n > 2$ we can find $n$ distinct positive integers such that the sum of their reciprocals is $1$ .

Solution

We have $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$ .

Now we have $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^n} + \frac{1}{2^n} = 1$ , replace the second $\frac{1}{2^n}$ by $\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{6}\right) \frac{1}{2^n}$ to get $n + 3$ terms ( $n \ge 1$ ):

$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^{n+1}} + \frac{1}{3 \cdot 2^n} + \frac{1}{3 \cdot 2^{n+1}} = 1$ .

For example, we have $\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{12} = 1$ and $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24} = 1$ .

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