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A circle is divided into six sectors. Then the numbers $1, 0, 1, 0, 0, 0$ are written into the sectors (counter-clockwise say). You may increase two neighboring numbers by $1$. Is it possible to equalize all numbers by a sequence of such steps?

So, $a_1 = 1, a_2 = 0, a_3 = 1, a_{4, 5, 6} = 0$.

The invariant they find is:

$$S = a_1 - a_2 + a_3 - a_4 + a_5 - a_6 = 2$$

If you increase $a_k$ by $1$ then you have:

$a_{2, 4, 5, 6} = 1, a_{1, 3} = 2$

$$S = 2$$ Still.

Then he says, the goal, $S = 0$ cannot be reached.

How does that solve the issue?

$$S=0 \implies a_1 + a_3 + a_5 = a_2 + a_4 + a_6$$

How does that do anything?

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Your target is to have all six numbers the same, i.e. $$a_1 = a_3 = a_5 = a_2 = a_4 = a_6$$ which would then imply $$a_1 + a_3 + a_5 = a_2 + a_4 + a_6$$ and so $$S=0$$ which we know is impossible from the starting point as $S$ is a constant $2$ under the permitted change.

The argument in words is that if the sum of the values in the even positions and the sum of the values in the odd positions start as being different, and all you can do is add the same numbers to odd and to even positions, then the sums will remain different, and therefore the individual values cannot all be made the same.

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  • $\begingroup$ I am trying to see the implication. $$a_1 = a_3 = a_5 = a_2 = a_4 = a_6$$ $$a_1 = a_2$$ $$a_3 = a_4$$ $$a_5 = a_6$$ $\sum \implies a_1 + a_3 + a_5 = a_2 + a_4 + a_6$. Got it. $\endgroup$ – Lebes Apr 23 '15 at 14:33
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I fail to see how $a_1−a_2+a_3−a_4+a_5−a_6$ is an invariant. Additionally I think I found a way to equalize the sectors. It is as follows:

$(1, 0, 1, 0, 0, 0) \rightarrow (1, 1, 1, 0, 0, 1) \rightarrow (2, 1, 1, 0, 1, 1) \rightarrow (2, 2, 1, 1, 1, 1)\rightarrow (2, 2, 1, 2, 1, 2) \rightarrow (2, 2, 2, 2, 2, 2)$

Where the tuples of numbers are the values in the circular sectors written counter clockwise. It's possible I'm miss understanding the problem though.

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$1,2,3,4,5,6$ are name of sectors counter-clock wise having $(1,0,1,0,0,0)$ initially. suppose the number of addition in $(1,2), (2,3), (3,4), (4,5), (5,6)$ and $(6,1)$ are $a,b,c,d,e$ and $f$ respectively.

If all numbers are equal, then $$a+f+1 = a+b = b+c+1 = d+c = d+e = e+f$$

which means $$a = c+1 , b = f+1, c = e , d = f , b+1 = d$$ we express $b$ in terms of $d$ we get $$b =f+1 = d+1$$ and we also have $$b+1 = d$$ hence
$$(d+1) + 1 =d$$ $$d+2 =d$$ which is not possible.

hence it is not possible to equalize all numbers by a sequence of such steps.

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