riddle of the day

A problem of transfers and invariants disguised as buckets and marbles: the move looks local, but the decisive idea is global.

The Last Bucket Full

Strategist
Numerical territory

Riddle statement

There are three buckets with marbles.

In each move, you may choose one bucket and double its number of marbles, taking exactly that many marbles from the other two buckets combined.

For example, if the buckets contain 5, 4 and 3 marbles, one way to double the bucket with 3 marbles is to take 2 from the bucket with 5 and 1 from the bucket with 4. You would get 3, 3 and 6.

Now compare these two cases:

  • Case A: 12, 7 and 3 marbles.
  • Case B: 10, 8 and 5 marbles.

In each case, can you make two buckets empty?

Show solution

Solution

Answer: Case A is possible; Case B is impossible.

In Case A there are:

$ 12+7+3=22 $

marbles. We can make all of them end up in a single bucket in two moves.

First double the bucket with 3 marbles. To do that, take 1 marble from the bucket with 12 and 2 marbles from the bucket with 7:

$ 12,\ 7,\ 3 \longrightarrow 11,\ 5,\ 6. $

Now the bucket with 11 marbles has exactly as many marbles as the other two buckets combined:

$ 5+6=11. $

Double that bucket with 11 marbles by taking all the marbles from the other two:

$ 11,\ 5,\ 6 \longrightarrow 22,\ 0,\ 0. $

So Case A is possible.

In Case B, however, the total is:

$ 10+8+5=23. $

To finish with two empty buckets, the final move would have to turn a position like this:

$ x,\ x,\ 0 $

into this:

$ 2x,\ 0,\ 0. $

In other words, just before the final move, half the marbles would have to be in the bucket being doubled, and the other half in the other buckets.

But if the total is 23, there is no whole-number half. You cannot split 23 marbles into two equal groups.

That is why Case B is impossible.