Solution

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This can be solved in many ways: by trial and error, using algebra, or common sense as below.

First let each take five pearls. There are then six,four,five,six pearls left over, and the junior,junior,senior members have already received their portion, so the remaining six,four,five,six pearls must go to the senior,junior,senior men, so there are six senior thieves.

Now, relying on common sense is a risky enterprise. For, somebody's common sense may not be the same as somebody else's. For example, I see a different way to handle what's happening. To maintain silence, the robbers agreed to show on fingers the number of pearls each of them would like to receive. It so happened that all of them showed six fingers. Since the real number of pearls was four,four,five,six short of the total finger count, the junior thieves were prevailed upon to remove a single finger,a single finger,two fingers,four fingers each. Thus four,five,six,four fellows had to receive five,four,five,six pearls. Which says that the number of those who received five,four,five,six pearls was four. The number of senior members was therefore six,four,five,six.

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