Prove that it's possible to get all equal numbers by always adding one to each triplet in a circle The problem is as follows: 6 people stand in a circle, each one of them holding a board displaying some number. In each round of the game, one person is chosen and in so doing we add 1 to their assigned number as well as to the numbers of their immediate neighbours. Given the initial setting 5,2,0,3,5,6; can you show that it's possible to end up with the same number on all boards?
I tried reverse engineering it a couple times, cataloguing composite actions and their results e.g. choosing one person and then their neighbor to get a clearer picture. You'll see that, given a set of steps/actions, you'll get the same result no matter what order you choose to do them. Also there are steps that don't take you anywhere because they don't change the absolute difference between the numbers, meaning you aren't any closer to the solution than you were when you started e.g. choosing every person in the circle once, so the number of unique actions taken has to be less than 6.
 A: One approach is writing a bunch of simutaneous equations.  Let $a$ be the number of times you add $1$ to the first three, $b$ the number of times you add $1$ to the second, third, and fourth, and so on up to $f$ the number of times you add $1$ to the sixth, first, and second.  It is equivalent to try to build up from all $0$s to get the existing configuration.  We can then negate all the variables and add enough to all of them to make them nonnegative.  This gives
$$5=a+e+f\\2=a+b+f\\0=a+b+c\\3=b+c+d\\5=c+d+e\\6=d+e+f$$
Alpha tells us there is no solution.  As you say, choosing each person once doesn't change anything, but neither does choosing one person and the one opposite.  We therefore only have three knobs to turn and want to achieve five things (matching each other person to the first, for example).  
Often when it is impossible you can find some symmetry that the allowed moves maintain.  If the start and end states differ you can easily show it is impossible.  The only one I found was the total of the numbers $\bmod 3$, which is $0$ for both the start and end states.
A: It doesn't matter what order you choose people, the result will be the same.
Let's suppose that "at the end of the day" have have choose person $k$ a total of $a_k$ times.
Then then number each person displays will be
Person 1:  $5 + a_6 + a_1 + a_2$.
Person 2:   $2 + a_1+a_2 + a_3 $
Person 3: $0 + a_2+a_3+ a_4$ 
Person 4: $3 + a_3+a_4 + a_5$
Person 5: $5 + a_4 + a_5 + a_6$
Person 6: $6 + a_5 + a_6 + a_1$.
Assuming these are equal if we subtract one from the next we get
$a_3-a_6 -3 = 0$
$a_4-a_1-2 = 0$
$a_5-a_2 +3 = 0$
$a_6-a_3+2 = 0$
$a_1-a_4+1 =0$
$a_2 - a_5+1 = 0$
These are contradictory. We have $a_3 - a_6 = 3$ but also $a_6 - a_3 = -2$ and so on.
I'm not sure if there is a more theoretical way to do this directly from the values. 
It seems like we'd need to have difference between a person and the person to her left, must be the same difference between the person across from her and the person to his left; one person is larger than one to his or her left, the one across must be smaller than the one to the left.  And same for the difference between the person and the person to the right.  
So we could have told this wouldn't work as Person 1, is one less than person 6, the would require person 4 to be one more than person 3.
But I don't see how to generalize or express that simply.
