Bariz - Problem

Source: 2022 Peru MO (ONEM) L3 p4 - finals

Tags: number theory, Perfect Powers, Perfect power, Digits

parmenides51

11.05.2024 23:45

For each positive integer n, the number $R(n) = 11 ... 1$ is defined, which is made up of exactly $n$ digits equal to $1$. For example, $R(5) = 11111$. Let $n > 4$ be an integer for which, by writing all the positive divisors of $R(n)$, it is true that each written digit belongs to the set $\{0, 1\}$. Show that $n$ is a power of an odd prime number. Clarification: A power of an odd prime number is a number of the form $p^a$, where $p$ is an odd prime number and $a$ is a positive integer.

1. Let $n=2^kt$ where $t$ is a odd number $\dfrac{10^{2^kt}-1}{9}=\dfrac{10^t-1}{9}(10^t+1)(10^{2t}+1)(10^{4t}+1)\ldots$ If $t=1$ then $10^4+1=73\cdot 137$ If $t>1$ $11 \mid 10^t+1$ $100\ldots 01:11=9\ldots$ 2. If $ 3 \mid n$ then $3 \mid R(n)$ 3. If $n=pq$ is a odd number where $\gcd(p,q)=1, p>1,q>1$ then: $\dfrac{10^p-1}{9} \mid R(n)$ $\dfrac{10^q-1}{9} \mid R(n)$ and $\gcd\left(\dfrac{10^p-1}{9},\dfrac{10^q-1}{9}\right)=1$ It is easy to prove that $\gcd(10^p-1,10^q-1)=10^{\gcd(p,q)}-1$ $\Longrightarrow \dfrac{10^p-1}{9}\cdot \dfrac{10^q-1}{9} \mid R(n)$ $\ldots \equiv 11 \cdot 11 \pmod{100} \equiv 21 \pmod{100}$ So, $n$ is a power of an odd prime.