On the board there are written in a row, the integers from $1$ until $2030$ (included that) in an increasing order. We have the right of ''movement'' $K$: We choose any two numbers $a,b$ that are written in consecutive positions and we replace the pair $(a,b)$ by the number $(a-b)^{2020}$. We repeat the movement $K$, many times until only one number remains written on the board. Determine whether it would be possible, that number to be: (i) $2020^{2020}$ (ii)$2021^{2020}$
Problem
Source: Greek National 2020 p3 (Archimedes)
Tags: combinatorics, number theory
VicKmath7
23.02.2020 22:54
The parity of the number of odd numbers on the board is invariant. So the last number is odd.
silouan
03.03.2020 17:20
Bump! Bump!
Jalil_Huseynov
16.01.2022 20:33
Since $a+b\equiv (a-b)^{2020} \pmod{2}$ we get last number is equal $1+2+\cdots 2030 \pmod{2}$. So last number should be odd. So $2020^{2020}$ can't be last number. Now I will show that $2021^{2020}$ can be last number. Applying movement $K$ to pairs $(2023,2024), (2025,2026),\cdots (2029,2030)$ gives $...2022,1,1,1,1$ and obvioulsy we can get $0$ from $1,1,1,1$ with $K$ movements. So we can get $1,2,3,\cdots , 2022$. Applying movement $K$ to pairs $(2,3),(4,5),\cdots, (2020,2021)$ gives $1,1,1,\cdots, 1,2022$, where there are $1011$ times $1$. Again we can get $0$ from $1,1,\cdots,1(1010$ times $1)$ with $K$ movements. So we can get $0,1,2022 \implies 1,2022 \implies 2021^{2020}$.
Enthurelx
19.10.2022 16:51
Jalil_Huseynov wrote: Since $a+b\equiv (a-b)^{2020} \pmod{2}$ we get last number is equal $1+2+\cdots 2030 \pmod{2}$. So last number should be odd. So $2020^{2020}$ can't be last number. Now I will show that $2021^{2020}$ can be last number. Applying movement $K$ to pairs $(2023,2024), (2025,2026),\cdots (2029,2030)$ gives $...2022,1,1,1,1$ and obvioulsy we can get $0$ from $1,1,1,1$ with $K$ movements. So we can get $1,2,3,\cdots , 2022$. Applying movement $K$ to pairs $(2,3),(4,5),\cdots, (2020,2021)$ gives $1,1,1,\cdots, 1,2022$, where there are $1011$ times $1$. Again we can get $0$ from $1,1,\cdots,1(1010$ times $1)$ with $K$ movements. So we can get $0,1,2022 \implies 1,2022 \implies 2021^{2020}$. It's actually a fake solve. You don't get $0,1,2022$ you get $1,0,2022,0$ which derives $1,2022,0$ you can't shuffle the sequence.