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 Monkeys and Coconuts (Posted on 2003-03-26)
Five men were stuck in a desert island. The island was plentifully supplied with coconuts, fortunately, and they spent the first day gathering the coconuts, heaping them in a great pile.

That night, one of the men was sleepless at the thought that everyone would fight over the coconuts at daybreak. So he arose, crept to the pile, and counted out the coconuts. Except for one coconut, it was possible to divide the pile into five equal portions. He gave the extra coconut to the monkey (who was the only other living being on the island besides those five men), buried his share of the coconuts, then went back to a sound sleep leaving the rest again in a pile.

Then another man got up and stole to the pile with the same object in mind. Again there was one coconut left over. The second man gave the spare coconut to the monkey, buried (what he thought was) his fair share, and went back to sleep, again leaving the remaining coconuts in one pile.

One by one, the remaining men got up and did the same and everytime the monkey got his share of one coconut.

In the morning, the men divided whatever coconuts remained into five equal piles, each taking his share and saying nothing about the night just past.

What is the minimum number of coconuts the five men assembled on the first day, if in the Final Division:

(a)There was again one coconut left over and the men (as usual) gave it to the monkey.

(b) There was no coconut left over for the monkey.

 Submitted by Ravi Raja Rating: 3.8571 (7 votes) Solution: (Hide) Let initially there be ‘x’ coconuts in the pile. Now, we proceed step by step as follows: Number of coconuts before the First division = x Number of coconuts removed in the First division = [{(x - 1)/5} + 1] Remaining number of coconuts after the First division = [4(x - 1)/5] = A (say) Number of coconuts before the Second division = A Number of coconuts removed in the Second division = [{(A - 1)/5} + 1] Remaining number of coconuts after the Second division = [4(A - 1)/5] = B (say) Number of coconuts before the Third division = B Number of coconuts removed in the Third division = [{(B - 1)/5} + 1] Remaining number of coconuts after the Third division = [4(B - 1)/5] = C (say) Number of coconuts before the Fourth division = C Number of coconuts removed in the Fourth division = [{(C - 1)/5} + 1] Remaining number of coconuts after the Fourth division = [4(C - 1)/5] = D (say) Number of coconuts before the Fifth division = D Number of coconuts removed in the Fifth division = [{(D - 1)/5} + 1] Remaining number of coconuts after the Fifth division = [4(D - 1)/5] = E (say) Now, for the first part of the problem, the number of coconuts remaining (after the fifth man finished dealing with the pile) is one more than a multiple of 5, i.e., (a) 5y + 1 (say) whereas for the second part, i.e., (b) the number of coconuts remaining is an exact multiple of 5, i.e., 5y (say) where ‘y’ is the number of coconuts each man receives in the final division. On simplification (getting the final expression in terms of x and y), we get for the two questions (the two parts of the problem) the following equations: 1024x - 15625y = 11529 for (a) and 1024x - 15625y = 8404 for (b). From these equations, we obtain: x (minimum) = 15621 for (a) and x (minimum) = 3121 for (b). Thus the number of coconuts left after each of the men finished dealing with the pile are: 15621, 12496, 9996, 7996, 6396, 5116 for (a) and 3121, 2496, 1996, 1596, 1276, 1020 for (b).

 Subject Author Date answers Dej Mar 2009-10-24 17:15:32 A ingenious solution for the equation (official solution) pcbouhid 2005-06-28 14:36:36 re(2): Comment on official solution Nick Hobson 2005-06-01 21:41:10 re: Comment on official solution Ravi Raja 2005-05-25 06:48:38 Comment on official solution Nick Hobson 2004-03-04 17:46:09 wierd helena 2003-07-14 05:19:10 re(3): One-line Simple Solution Gamer 2003-07-02 06:04:44 re(2): One-line Simple Solution Jayaram S 2003-05-12 00:01:32 re: One-line Simple Solution Ravi Raja 2003-05-09 04:18:08 re(3): i got it! Ravi Raja 2003-05-09 04:10:10 One-line Simple Solution Jayaram S 2003-05-08 21:37:23 re(2): i got it! sendil 2003-05-07 21:21:30 ANSWER Chaz 2003-05-03 04:14:22 re: i got it! Ravi Raja 2003-04-03 03:41:32 i got it! Manoj Nair 2003-04-03 03:21:56 re: possible sol Ravi Raja 2003-03-31 23:42:14 possible sol hj 2003-03-31 11:42:38 re: Possible solution Ravi Raja 2003-03-28 22:37:13 re: Possible Solution Ravi Raja 2003-03-28 22:32:08 OOPS! Possible solution Trevor John Streeton 2003-03-28 21:44:49 Possible solution Trevor John Streeton 2003-03-28 21:28:29 Possible Solution isaiah 2003-03-27 14:14:42 re(5): Possible solution - smaller solution Charlie 2003-03-27 09:28:47 re(4): Possible solution - smaller solution Charlie 2003-03-27 09:27:06 re: You don't wanna know... fwaff 2003-03-27 06:12:04 You don't wanna know... jude 2003-03-27 04:16:42 re: you two are right Ravi Raja 2003-03-27 03:14:09 re(2): Possible solution - smaller solution Ravi Raja 2003-03-27 03:11:00 re(3): Possible solution - smaller solution Ravi Raja 2003-03-27 03:08:05 re: Possible solution Ravi Raja 2003-03-27 03:04:33 re: Solution part 1 Bryan 2003-03-26 11:15:47 Solution part 2 Fernando 2003-03-26 09:43:15 Solution part 1 Fernando 2003-03-26 09:41:59 you two are right jude 2003-03-26 07:18:36 re(2): Possible solution - smaller solution Bryan 2003-03-26 07:00:07 re: Possible solution - smaller solution fwaff 2003-03-26 05:40:09 Possible solution jude 2003-03-26 05:26:05

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