Five men collected few number of coconuts. They decided to divide the coconuts into 5 equal groups next day...

Question

That night each man took a turn watching for rescue searchers while the others slept. First watcher got bored so he decided to divide the coconuts into 5 equal piles. When he did this he found he had 1 remaining coconut, he gave that coconut to monkey, took 1 of the piles and hid it for himself then he jumbled up the 4 other piles into 1 big pile again.
In short each of the five men ended up doing exactly the same thing.
What is the smallest number of coconuts there could have been in the original pile.

Answer 1

Let x = the number of coconuts at the beginning

Then there were 4(x-1)/5 after the first man finished his watch.

There were 4(4(x-1)/5-1)/5 after the second man finished his watch.
= (16x-36)/25

There were 4(4(4(x-1)/5-1)/5-1)/5 after the third.
= (64x - 244)/125

There were 4(4(4(4(x-1)/5-1)/5-1)/5-1)/5 after the forth.
= (256x - 1476)/625

And after the fifth man finished, there were:
4(4(4(4(4(x-1)/5-1)/5-1)/5-1)/5-1)/5
= (1024x - 8404)/3125

This means that (1024x – 8404) must be divisible by 3125 (since there were never any fractions of coconuts!)

I set up a spread sheet (I had no intention of doing that math by hand!) and Nghiem E is correct -- 3121 is the lowest number of coconuts the men could have originally collected.



They then started with 3121, the first man hid 624 and gave 1 to the monkey, leaving 2496.

The second man hid 499 and gave 1 to the monkey, leaving 1996.

The third man hid 399 and gave 1 to the monkey, leaving 1596.

The fourth man hid 319 and gave 1 to the monkey, leaving 1276.

The fifth and last man hid 255, and gave 1 to the monkey leaving 1020 coconuts.

I hope this all makes sense!

Answer 2

Every sailor leaves 4/5(n-1) coconuts of a pile of n coconuts. This results in an awful formula for the complete process (because every time one coconut must be taken away to make the pile divisible by 5):

4/5(4/5(4/5(4/5(4/5(4/5(p-1)-1)-1)-1)-1)-1), where p is the number of coconuts in the original pile, must be a whole number.

The trick is to make the number of coconuts in the pile divisible by 5, by adding 4 coconuts. This is possible because you can take away those 4 coconuts again after taking away one fifth part of the pile: normally, 4/5(n-1) coconuts are left of a pile of n coconuts; now 4/5(n+4)=4/5(n-1)+4 coconuts are left of a pile of n+4 coconuts. Because of this, the number of coconuts in the pile stays divisible by 5 during the whole process. So we are now looking for a p for which the following holds:

4/5×4/5×4/5×4/5×4/5×4/5×(p+4)=(46/56)×(p+4), where p is the number of coconuts in the original pile, must be a whole number.

The smallest (p+4) for which the above holds, is 56. So there were p=56-4=15621 coconuts in the original pile.

Answer 3

(a) After the fifth man is done, if the remainder has to be divided evenly by 5, then the answer is "3121"
according to:
http://acm.uva.es/p/v6/616.html (copied below)

(b) Otherwise, if it doesn't matter what is left after the fifth man is done, "1871" will work.

I made a chart using
4x (after the fifth man is done)
5x+1 = 4a (after the fourth man is done)
5a+1 = 4b (after the third man is done)
5b+1 = 4c (after the second man is done)
5c+1 = 4d (after the first man is done)
5d + 1 (starting number of coconuts)

Values of x that make 5x + 1 a multiple of 4
x = 3, 7, 11, 15, 19, 23, (adding 4 each time)
Every fourth number in this sequence can
make the next level also a multiple of 4
x = 15, 31, 47, 63 (works for 4 men but not 5),
79, 95, 111, 127 (works for 4 men but not 5),
143, 159, 175, 191 (works for 5 men)

[191 x 5 + 1] / 4 = 239
[239 x 5 + 1] / 4 = 299
[299 x 5 + 1] / 4 = 374

So 374 x 5 + 1 = 1877 starting number of coconuts

(a) If the remainder has to be divisible by 5
the smallest number is
"3121"
according to:

"Coconuts, Revisited"

The short story titled Coconuts, by Ben Ames Williams, appeared in the Saturday Evening Post on October 9, 1926. The story tells about five men and a monkey who were shipwrecked on an island. They spent the first night gathering coconuts. During the night, one man woke up and decided to take his share of the coconuts. He divided them into five piles. One coconut was left over so he gave it to the monkey, then hid his share and went back to sleep.

Soon a second man woke up and did the same thing. After dividing the coconuts into five piles, one coconut was left over which he gave to the monkey. He then hid his share and went back to bed. The third, fourth, and fifth man followed exactly the same procedure. The next morning, after they all woke up, they divided the remaining coconuts into five equal shares. This time no coconuts were left over.

An obvious question is ``how many coconuts did they originally gather?" There are an infinite number of answers, but the lowest of these is 3,121.

Answer 4

if each person removes 2 coconuts, then by the end of the night 10 are removed.
therefore the smallest number of coconuts there could have been is 10