An equilateral triangle with side length $3$ is divided into $9$ congruent triangular cells as shown in the figure below. Initially all the cells contain $0$. A move consists of selecting two adjacent cells (i.e., cells sharing a common boundary) and either increasing or decreasing the numbers in both the cells by $1$ simultaneously. Determine all positive integers $n$ such that after performing several such moves one can obtain $9$ consecutive numbers $n,(n+1),\cdots ,(n+8)$ in some order. [asy][asy] size(3cm); pair A=(0,0),D=(1,0),B,C,E,F,G,H,I; G=rotate(60,A)*D; B=(1/3)*D; C=2*B;I=(1/3)*G;H=2*I;E=C+I-A;F=H+B-A; draw(A--D--G--A^^B--F--H--C--E--I--B,black);[/asy][/asy]
Problem
Source: India IMO Training Camp 2016, Practice 2, Problem 3
Tags: combinatorics
SidVicious
22.07.2016 12:43
It's impossible for $n=1$ because of parity of sum. For $n=2$ it can be easily obtained. Suppose it can be done for $n>2$. Now mark the little triangles with $1,2,...,9$ ($1$ is on the left bottom and $5$ is on the right bottom). Color $2,4,7$ in black and let $S_{2,4,7}$ be the sum of numbers in $2,4,7$. Similarly define $S_{1,3,5,6,8,9}$. Obviously in any moment we have $S_{2,4,7}=S_{1,3,5,6,8,9}$ so it also holds when the table is filled with $n,(n+1),...,(n+8)$. But $S_{2,4,7} \le (n+8)+(n+7)+(n+6)=3n+21$ and $S_{1,3,5,6,8,9} \ge n+...+(n+5)=6n+15 > 3n+21 \ge S_{2,4,7}$ so $S_{2,4,7} < S_{1,3,5,6,8,9}$ which is not true. So the answer is $n=2$.
galav
01.04.2017 05:40
How to do for $n=2$? I got one way after trying out many. any method?
WizardMath
01.04.2017 07:29
@above, look at the equality cases and systematically add one to each pair of triangles, a way to do this would be first doing this operation on corner and make your way to the centroid. My way was to consider the max and min $f$ where $f$ is defined as follows: Do a chessboard coloring then $f= $ sum of black - sum of white.
Chikara
23.12.2024 20:05
We color the cells black and white alternately where two adjacent cells are not the same color (assume the corners are colored white). Let $S$ be the sum of numbers in cells. Now it is Obvious that $S_{B}=S_{W}$. We also know that $S_{B}+S_{W}=n+(n+1)+...+(n+8)=9n+36$. Therefore $S_{B}=S_{W}=\frac{9}{2}n+18$ (hence proving $n\in E$). Because $S_{B}$ is equal to sum of only 3 numbers, $\frac{9}{2}n+18\le (n+6)+(n+7)+(n+8)=3n+21$. Thus $n\le 2$ and the only solution is $n=2$.