A king was hungry one night and went downstairs to get some food ... He saw mangos in a dish and took 1/6 of them ... Then his wife went downstairs and got 1/5 of the remaining mangos ... Then his son came down and took 1/4 of the remaining mangos ... Then his oldest son went down and took 1/3 of the remaining mangos ... Last his youngest son came down and took 1/2 of the mangos left ... There are 3 mangos left ... How many mangos were in the dish at the beginning?

Hint 1

Work backward from the number of mangos left at the end.

Hint 2

The youngest son took 1/2 of the mangos he saw, leaving 1/2 of the ones he saw, or 3 mangos. Therefore the number he saw was 3 / (1/2) = 6 mangos. Keep working backward.

Complete Solution

The youngest son took 1/2 of the mangos he saw, leaving 1/2 of the ones he saw, or 3 mangos. Therefore the number he saw was 3 / (1/2) = 6 mangos.
The oldest son took 1/3 of the mangos he saw, leaving 2/3 of the ones he saw, or 6 mangos. Therefore the number he saw was 6 / (2/3) = 9 mangos.
The first son took 1/4 of the mangos he saw, leaving 3/4 of the ones he saw, or 9 mangos. Therefore the number he saw was 9 / (3/4) = 12 mangos.
The wife took 1/5 of the mangos she saw, leaving 4/5 of the ones she saw, or 12 mangos. Therefore the number she saw was 12 / (4/5) = 15 mangos.
The king took 1/6 of the mangos he saw, leaving 5/6 of the ones he saw, or 15 mangos. Therefore the number he saw was 15 / (5/6) = 18 mangos.

Check your work, working forward:

The king took 1/6 of 18, leaving 15
The wife took 1/5 of 15, leaving 12
The first son took 1/4 of 12, leaving 9
The oldest son took 1/3 of 9, leaving 6
The youngest son took 1/3 of 6, leaving 3

Working Backward

In this problem we started with an unknown number of mangos, took away some other unknown numbers of mangos, until we finally had 3 mangos left. To solve the problem we worked backward: We started with the known number of mangos at the end and worked backward in time until we got to a definite number of mangos at the start.

If you know algebra you might have worked the problem like this: Let M be the number of mangos at the beginning. The king takes 1/6 of these, or (1/6) M, leaving (5/6) M. The wife takes 1/5 of these, leaving (4/5)(5/6)M = (4/6)M. We continue like this, until we conclude that the youngest son left (1/6)M. Since we also know that this is 3 mangos, we know (1/6)M = 3, which we can solve as M = 18. In a way this is working backward too; we hold the original amount unknown, and finally "work backward" from (1/6)M = 3 to M = 18. This works algebraically because the very simple relationship between each step allows us to simplify as we go along and get an algebraic equation that we can solve.

Pólya gives another example: If you need to fetch exactly 6 quarts of water from the river, and you have only a 9 quart pail and a 4 quart pail, neither having any volume markings on the side, how can you do it? By working backward: There will be several steps involved, and after the last step you will have 6 quarts in the 9 quart pail. What would be in the pails just before this step? And what would be in the pails the step before that?

References

George Pólya, How To Solve It, Princeton University Press, 1957, pp. 225-232.
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