mathematics - Nine gangsters and a gold bar

One night nine gangsters stole a gold bar. When the time came for dividing the bar, they faced a problem: two of the criminals put guns to each other's faces. Now it's up to fate whether one of them lives, they both live or both die.

While these two are dealing with each other, the others decide to continue dividing the gold bar. What is the minimal amount of pieces they should divide the bar into, so that no matter how things pan out, everyone can be given an equal share?

Scenario 1: Both gangsters blow each other's brains out. The gold must be divided evenly among the seven remaining gangsters.

Scenario 2: One gangster is quicker on the draw, and manages to take out his opponent. The gold must be divided evenly among the eight remaining gangsters.

Scenario 3: The duelling gangsters discuss their differences, come to a mutually beneficial agreement, and put away their guns. The gold must be divided evenly among all nine gangsters.

Answer

18 pieces:

3,7,9,14,16,18,23,24,25,30,31,32,38,40,42,47,49,56

In 9 parts:

{3,23,30} {7,49} {9,47} {14,42} {16,40} {18,38} {24,32} {25,31} {56}

In 8 parts:

{3,18,42} {7,56} {9,24,30} {14,49} {16,47} {23,40} {25,38} {31,32}

In 7 parts:

{3,7,24,38} {9,14,18,31} {16,56} {23,49} {25,47} {30,42} {32,40}

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We need to be able to split it into 9 parts of 56, so it can't hurt to make 9 pieces of 56 and then split those further. Since we need to do better than 19 pieces, we can have at most 18 pieces. This means that most of our 56s are split into exactly two parts (we can have an extra piece for every 56 we don't split).

Now, what will our pieces be mod 9? We must be able to group them into 7 or 8 groups that sum to 0, or make 9 pairs (mostly) that sum to 2. If we have a 2, we can group it with a 7 to make 0. Then we pair the 7 with a 4, and repeat with a 5 and 6. If we do this 3 times, we can use the 6s to make another 0.

To make 63s and 72s, we should repeat at an interval of 9, so that we can swap the pairs to add 9 (e.g. {7,56} {16,47} {25,38} -> {16,56} {25,47}). The other two numbers that are swapped out need an extra 27.

Suppose we start with a 56 (freeing up one cut) and see what numbers we need.

    56-> 7->49->14->42
 9->47->16->40->23->33
18->38->25->31->32->24

So far, this makes 6 63s with 9 18 24 33 42 left over. We also need to make a 27 and a 72 out of 24 33 42. If we split 33 into 3 and 30, both of these are resolved. This results in the final answer with 18 pieces.