# Thread: a truly ill and confusing math problem

1. ## a truly ill and confusing math problem

On a deserted island live five people and a monkey. One day everybody gathers coconuts and puts them together in a community pile, to be divided the next day. During the night one person decides to take his share himself. He divides the coconuts into five equal piles, with one coconut left over. He gives the extra coconut to the monkey, hides his pile, and puts the other four piles back into a single pile. The other four islanders then do the same thing, one at a time, each giving one coconut to the monkey to make the piles divide equally. What is the smallest possible number of coconuts in the original pile?

2. 3906 ?

Ammo

3. nope

4. They keep splitting the pile in 5 right (like they don't know they others already took their part)?.

Hum.. I think I missed an iteration...: 19531?

Ammo

5. this problem is an old Russian Olympiad problem (Gr. 7/8 olympiad)
also seems to be the beginning of CRT (chinese remainder theorem)
careful reading does the trick. j/k

The answer is a solution to a Diophantine equation (i.e., equations requiring integral solutions to unknowns) and is rather a large number.

The original pile must be a number such that you can subtract one and multiply by 4/5 and get an integer. Possibilities are 6, 11, 16, 21, 26 ...
After the first person is finished the remaining pile will have 4, 8, 12, 16, 20, ...

The remaining pile after the first person must be a number such that you can subtract one then multiply by 4/5 and get an integer. Possibilities are 16, 36, 56, 76, 96, ...

After the second person is finished the remaining pile will have 12, 28, 44, 60, 76, ...

The remaining pile after the second person must be a number such that you can subtract one then multiply by 4/5 and get an integer. Possibilities are 76, 156, 236, 316, 396, ...

After the third person is finished the remaining pile will have 60, 124, 188, 252, 316 ...

The remaining pile after the third person must be a number such that you can subtract one then multiply by 4/5 and get an integer. Possibilities are 316, 636, 956, 1276, 1596, ...

After the fourth person is finished the remaining pile will have 252, 508, 764, 1020, 1276 ...

The remaining pile after the fourth person must be a number such that you can subtract one then multiply by 4/5 and get an integer. The smallest possibilities is 1276

After the fifth person is finished the remaining pile will have 1020 ...

So the fifth person will get 1276-1020-1 = 255.

The fifth person will leave behind a pile of 1020.

The fourth person will have left behind a pile of 1020+255+1 = 1276.

The fourth person will get 1276/4 = 319.

The third person will have left behind a pile of 1276+319+1 = 1596.

The third person will get 1596/4 = 399.

The second person will have left behind a pile of 399+1596+1 = 1996.

The second person will get 1996/4 = 499.

The first person will have left behind a pile of 499+1996+1 = 2496.

The first person will get 2496/4 = 624.

The original pile must have had 624+2496+1 = 3121.

6. Good job, although it may have been more prudent to mark your post as hidden.

7. my apologies...first post. i'll get it right next time.

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