Of Monkeys and Men (and Coconuts)

Number Theory Level 3

Five men are trapped on an island and have collected a large number of coconuts for food. Before they fall asleep, they decide that they will take turns having one man on watch duty while the others sleep, in case a rescue helicopter comes by.

Once the other men have fallen asleep, the first man divides the pile into five equal groups, giving the one leftover coconut to a curious monkey nearby. He then takes one of the five equal groups for himself, piling up the remaining coconuts. Finally, he wakes up the second man and goes to sleep.

The second, third, fourth, and fifth men repeat this process when it is their turn for watch duty, dividing the pile into fifths, giving the one leftover coconut to the observing monkey, and taking one of the fifths for themselves. The next morning, the men divide up the pile into five equal groups, with no remainder.

What is the smallest number of coconuts the men could have collected originally?

The answer is 3121.

1 solution

Nathan Ramesh
Dec 19, 2014

Neat. Use base 5. Obviously, the number in base 5 ends in $1$ , since when $1$ is subtracted, there is a multiple of $5$ . Think about the entire number $a_1a_2a_3a_4a_51_5$ . (We can assume it's $5$ digits since every digit added to the left of one guarantees us something, we will see later. In base $5$ , the second person will receive $a_1a_2a_3a_4 a_5 0-a_1a_2a_3a_4a_5$ . We know this number must end in a $1$ , thus $a_5=4$ . Now we have $a_1a_2a_3a_441_5$ . Using the same technique as before, we can solve for $a_1,a_2,a_3,a_4$ and get $0,4,4,4$ (the $4$ s are all the same for a very specific reason, try and see why). Thus the number is $44441_5=3121$

This solution also provides more insight than a standard algebraic approach. If there are $10$ men, for example, the answer would just be $4444444441_5$ .

Of Monkeys and Men (and Coconuts)

The answer is 3121.

1 solution

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