A puzzle question about mango thieves.

One night three naughty boys stole a basketful of mangoes from a garden, hid the loot and went to sleep. Before retiring they did some quick counting and found that the fruits were less than a hundred in number.

During the night one boy awoke, counted the mangoes and found that he could divide the mangoes into three equal parts if he first took one for himself. He then took one mango, ate it up, and took 1/3 of the rest, hid them seperately and went back to sleep.

Shortly thereafter another boy awoke, counted the mangoes and he again found that if he took one for himself the loot could be divided into three equal parts. He ate up one mango, bagged 1/3 of the remainder, hid them separately and went back to sleep. The third boy also awoke after some time, did the same and went back to sleep.

In the morning when they all awoke up, and counted their mangoes, they found that the remaining mangoes again totalled 1 more than could be divided into three equal parts.

How many mangoes did the boys steal?

Image via Wikipedia

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Assuming only 4 mangoes remained in the morning, this would mean that the third boy must have found 7 mangoes left when he woke up during night. But 7 is not 2/3 of a whole number, so this is impossible.

The next possibility is 7 mangoes left in the morning, which is again impossible.

Now the next possibity is 10, which is 2/3 of 15. This means that the third boy found 16 mangoes, took one and then took 5 more. The second boy then taken 8 more. But 25 is not 2/3 of a whole number and, therefore, the assumption that 10 mangoes remained in the morning is absurd.

By similar reasoning the numbers 13, 16 and 19 can be eliminated, but 22 will be found to meet the required conditions.

The third boy found 34, took one and left 2/3 of 33 or 22, the second boy found 52, took one and left 2/3 of 51 or 34, the first boy found 79 took 1 and left 2/3 of 78 or 52.

The answer is the boys stole altogether 79 mangoes.

Image via Wikipedia

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