Odd Stones
Problem
$27$ stones are distributed between $3$ circles
On a "move" a stone is removed from two of the circles and placed in the third circle.
So, in the illustration, if a stone is removed from the $4$ and the $10$ circles and added to the $13$ circle, the new distribution would be $3$ - $9$ - $15$
Check you can turn $2$ - $8$ - $17$ into $3$ - $9$ - $15$ in two "moves"
Here are five of the ways that $27$ stones could be distributed between the three circles :
$6$ - $9$ - $12$
$3$ - $9$ - $15$
$4$ - $10$ - $13$
$4$ - $9$ - $14$
$2$ - $8$ - $17$
There is always some sequence of "moves" that will turn each distribution into any of the others - apart from one.
Identify the distribution that does not belong with the other four.
Can you be certain that this is actually impossible rather than just hard and so far unsuccessful?
Getting Started
Hint for Check Point :
Turn $2$ - $8$ - $17$ into $4$ - $7$ - $16$
then $4$ - $7$ - $16$ into $3$ - $9$ - $15$
Hint for proving the odd one is impossible :
This example isn't the same thing but might give you a clue about the kind of thinking to try.
In a $4$ circle problem and using $26$ stones the distribution $1$ - $4$ - $7$ - $14$ cannot be turned into $3$ - $5$ - $7$ - $11$
To understand why notice that in the first there are two odd and two even numbers while in the second the numbers are all odd.
On a "move" one value goes up by $3$ and the others go down be one.
What will happen to odd and to even numbers?
The Odd Stones problem isn't about odd or even numbers but a similar kind of thinking could be useful.
Good luck!
Student Solutions
Good start from Hamish in New Zealand
$2$ - $8$ - $17$ goes to $4$ - $7$ - $16$
and from there to $3$ - $9$ - $15$
Ruth from Manchester High School then looked for the odd one out
$4$ - $9$ - $14$ is the impossible arrangement.
Consider the numbers in each circle in modulo $3$.
Modulo $3$ means the remainder amount when you divide a number by three.
In the first arrangement ($6$ - $9$ - $12$) the modulo $3$ value of each pile is $0$.
On each move you take $1$ from $2$ of the piles and add $2$ to the third so the numbers which were all $0$ in modulo $3$ now all become $2$ in modulo $3$, and after that $1$ in modulo $3$, then finally $0$ again.
After that the cycle just repeats over and over again.
For four of the arrangements the initial numbers are all equal in modulo $3$ and whatever you choose as the next move they will stay equal in modulo $3$.
But $4$ - $9$ -$14$ is $1$ - $0$ - $2$ in modulo $3$ and so cannot turn into any of the other four arrangements or be reached from them.
Thanks Ruth. That way of looking at numbers using their modulo value seems like a powerful perspective.
Teachers' Resources
This could be a useful extension activity helping students to break away from too readily expecting odd or even to be the important characteristic. Odd or even-ness can be seen more generally as the remainder after a division by two, and this problem depends on remainders using a different divisor.
This context has more possibilities than the simple question posed in the problem. It is capable of building up into a rich dynamical system well within the scope of a Stage 4 student. The number of stones and more especially the number of circles are the key variables.