4. There are several (at least two) positive integers written along the circle. For any two neighboring integers one is either twice as big as the other or five times as big as the other. Can the sum of all these integers equal 2023 ? Sergey Dvoryaninov
Problem
Source: 45th International Tournament of Towns, Junior O-Level P4, Fall 2023
Tags: number theory
Tintarn
16.12.2023 20:39
The answer is no. In fact, we show that the sum has to be a multiple of $3$ (and clearly all multiples of $3$ are indeed possible by alternating between $1$ and $2$).
Indeed, if one of the numbers is divisible by $3$, then all are and hence also the sum.
On the other hand, if none of the numbers is divisible by $3$, then their residues mod $3$ must alternate between $1$ and $2$. In particular, the number of numbers around the circle must be even, and their sum a multiple of $3$.
Warideeb
09.12.2024 21:26
The answer is No. For the sake of contradiction we assume there are numbers $a_1,a_2,...,a_n$ such they are on the circle and $a_1+a_2+...+a_n=2023$ where $a_i$ has neighbors $a_{i-1}$ and $a_{i+1}$ $(a_0=a_{2023},a_{n+1}=a_1)$ \(Claim\)$1$: $a_i \equiv a_{i-1} \pmod{3}$ is impossible. (\prove:\) WLOG, $a_i=2a_{i-1}$ or $5a_{i-1} \equiv 2a_i \pmod{3}$ Which proves our claim. \(Claim\)$2$: $a_i+a_{i+1} \equiv 0 \pmod{3}$ It can be proved like \(Claim1\) Now suppose $n$ is even. Now as $a_1+a_2+a_3+...+a_n=2023$ $(a_1+a_2)+(a_3+a_4)+...+(a_{n-1}+a_n)=2023$ $0 \equiv 2023 \pmod{3}$ which is impossible. So $n$ must be odd. Now $(a_1+a_2)+...+(a_{n-2}+a_{n-1})+a_n=2023$ So $a_n \equiv 1 \pmod{3}$ Now we pair up like this $(a_n+a_1)+(a_2+a_3)+...+(a_{n-3}+a_{n-2})+a_{n-1} =2023$ So $a_{n-1} \equiv 1 \pmod{3}$ That implies $a_{n-1} \equiv a_n \pmod{3}$ \(Contradiction!\) So not possible. $\square$