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Differences on a board

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The numbers $1$ to $9$ are written on a board. A move erases any $2$ numbers and writes the absolute value of their difference. After $8$ moves a single number is left. What is the smallest value it can have?

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Hint

Look at the parity of the total.

Solution

Replacing $a$ and $b$ by $∣ a - b ∣$ changes the sum by $2 \times min (a, b)$ , an even number, so the parity of the sum never changes.
The starting sum is $1 + 2 + \dots + 9 = 45$ , which is odd, so the last number is odd and at least $1$ .
The value $1$ is reachable: replace $9$ and $8$ , $7$ and $6$ , $5$ and $4$ , $3$ and $2$ by their differences, which leaves $5$ copies of $1$ .
Then $2$ pairs of these give $0$ and $0$ , those give $0$ , and the last move on $0$ and $1$ leaves $1$ .
So the smallest possible value is $1$ .

Answer: $1$

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