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Entry 03 / Mathematical puzzle / Oct 11 · 1:38 PM EDT

The Vanishing Numbers

Nine integers are reduced to one. Can the last number be zero? An invariant survives every apparent disappearance.

Full text supplied in Numbers-game.rtf. Complete text from the supplied RTF file. Wording is preserved; typography and paragraph layout are adapted for reading on the website.

I decided to invent a little game, then investigate whether it can actually be won. 🎲

The Vanishing Numbers

Write the integers

1,2,3,4,5,6,7,8,9 1,2,3,4,5,6,7,8,9

on a piece of paper.

On each turn, choose any two numbers a,b, erase them, and write down |a-b|.

For example, the first move could transform

1,2,3,4,5,6,7,8,9 1,2,3,4,5,6,7,8,9

into

1,2,3,4,5,6,7,5. 1,2,3,4,5,6,7,5.

Continue until only one number remains.

Your goal is to make that final number zero.

An unexpected discovery

The game is impossible to win.

The reason is wonderfully simple: the sum of the numbers is initially

1+2+⋯+9=45. 1+2+\cdots+9=45.

Whenever we replace a,b with |a-b|, the sum decreases by

a+b−|a−b|=2min(a,b). a+b-|a-b|=2\min(a,b).

Thus the sum always changes by an even integer. It begins odd, so it must remain odd.

The final number therefore cannot be zero.

A stranger variation

What if we replace a,b by their sum or absolute difference, freely choosing which operation to perform?

Still impossible! Addition preserves the sum, while subtraction changes it by an even number.

But now suppose we add a third permissible operation:

(a,b)↦⌊a+b2⌋. (a,b)\longmapsto \left\lfloor\frac{a+b}{2}\right\rfloor.

Suddenly the parity obstruction disappears.

For example, repeatedly averaging the two smallest available numbers (rounding down) eventually turns the original collection into zero:

(1,2)→1,(1,3)→2,(2,4)→3,(3,5)→4,(4,6)→5,(5,7)→6,(6,8)→7,(7,9)→8. \begin{aligned} &(1,2)\to1,\quad (1,3)\to2,\quad (2,4)\to3,\\ &(3,5)\to4,\quad (4,6)\to5,\quad (5,7)\to6,\\ &(6,8)\to7,\quad (7,9)\to8. \end{aligned}

That particular strategy finishes at 8, not zero. But there is a much simpler winning sequence: first combine 1 and 2 by subtraction to create 1; then combine that 1 with 3 by averaging to create 2; and so on—although this approach also requires care.

In fact, a winning sequence can be constructed immediately by making a zero with (1,1), after first obtaining two ones from (2,3) and (4,5). Once zero is available, average it with each remaining positive integer, repeatedly rounding down until the values are small enough to eliminate.

The interesting part isn’t the game itself. It’s the transition between two worlds: one where a conserved quantity makes an outcome impossible, and another where a tiny change in the rules destroys the obstruction.

A whole theory can sometimes begin with such a tiny change.

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