Competition problem — Number theory

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Number theory Difficulty 6.1 National Olympiad Prove it Romania

An integer $n > 1$ will be called p-periodic if $\frac{1}{n}$ is a repeating decimal fraction, whose shortest period has length $p$ and begins immediately after the decimal point. For instance, $\frac{1}{9} = 0.111\dots$ is 1-periodic and $\frac{1}{11} = 0.090909\dots$ is 2-periodic.

a) Find all p-periodic numbers $n$ so that the first digit of the period of $\frac{1}{n}$ is not nil.

b) Find the largest 4-periodic prime.

Solution

a) The first digit of the period is at least 1 if and only if $\frac{1}{n} \ge \frac{1}{10}$ , that is $n \le 10$ . Since the prime factors of $n$ must be different from 2 and 5, it follows that $n \in \{3, 7, 9\}$ ; indeed, in this case $\frac{1}{3} = 0,(3)$ , $\frac{1}{7} = 0,(142857)$ and $\frac{1}{9} = 0,(1)$ .

b) Let $p$ be the largest 4-periodic prime; then $\frac{1}{p} = \frac{m}{9999}$ , with $1 \le m \le 9998$ . This yields $pm = 9999 = 3^2 \cdot 11 \cdot 101$ . So, the candidate for the largest such prime is $p = 101$ , which is acceptable, because $\frac{1}{101} = 0,(0099)$ .

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