Monday, April 26, 2010
The men with the coconuts and monkey
Five men crash-land their airplane on a deserted island in the South Pacific. On their first day they gather as many coconuts as they can find into one big pile. They decide that, since it is getting dark, they will wait until the next day to divide the coconuts.
That night each man took a turn watching for rescue searchers while the others slept. The first watcher got bored so he decided to divide the coconuts into five equal piles. When he did this, he found he had one remaining coconut. He gave this coconut to a monkey, took one of the piles, and hid it for himself. Then he jumbled up the four other piles into one big pile again.
To cut a long story short, each of the five men ended up doing exactly the same thing. They each divided the coconuts into five equal piles and had one extra coconut left over, which they gave to the monkey. They each took one of the five piles and hid those coconuts. They each came back and jumbled up the remaining four piles into one big pile.
What is the smallest number of coconuts there could have been in the original pile?
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In the original pile there could be 6 coconuts, or a multiple of 6. I know this because a man always divides us the coconut into 5 piles and there's one left ex: 6 into 5 groups= 1 coconut in each group and one for the monkey.
ReplyDeleteIris
There would be 6 coconuts in the original pile. I figured this out by carefully reading the question, and then concluding that 6 coconuts in the first pile would make sense, because one for each person (there are 5 people) and then one for the monkey adds up to six. It's really easy once you get the trick!
ReplyDeleteKatty
As far as I have gone (and I have not gone far), IT IS IMPOSSIBLE. If one person divides... let's say 36 coconuts. He divides them into 5 groups of 7, and 1 for the monkey. This leaves 25 for the next person. 25 is divisible by 5, and none for the monkey.
ReplyDeleteIf there were 31 coconuts to start with, the guy would divide them in 5 piles of 6, and 1 for the monkey. He takes one pile of 6, and there are 25 coconuts left, still divisible by 5.
So it keeps on going, they end up at a number divisible by 5. Unless I use bigger numbers...
Let's try 101 coconuts. He splits them into 5 portions of 20, and 1 for the monkey. He takes a pile, and puts the rest together. That leaves 80 coconuts left for the others, also divisible by 5.
How about an odd number...17. 17✕5=85. 85+1(for the monkey)=86. There are 86 coconuts to start with. The stranded man divides them into 17 groups of 5, and 1 for the monkey. He takes 17 and there are 68 coconuts left. Not divisible by 5! So, the next watcher comes and gets bored so he divides the 68 coconuts into 5 groups of 13, and 1 for the monkey. He realizes even after giving one to the monkey, there are 2 left. So he has no idea what to do with them and throws them in the ... water ... no not really.
I'm still working on this mind-bending puzzle.
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The smallest number i could find is 12496. First you do 12496-1=12495 divided by 5=2499. So you do 12495-2499=9996. Next step is 9996-1=9995 divided by 5=1999. So 9995-1999=7996. Then you do 7996-1=7995 divided by 5=1599 so 7955-1599=6396. Next step is 6396-1=6395 divided by 5=1279 so 6395-1279=5116. Next step is 5116-1=5115 divided by 5=1023 so 5115-1023=4092. That is 5 times repeated. This took me hours of guessing and equations.
ReplyDeleteUnless there is a trick, the question is impossible. Bu it seems like there is no hidden piece of information; each man made 5 piles, took one and gave a coconut to the monkey. If there were six coconuts then the first man would make five piles of one and give the last coconut to the monkey. But then the second man would only have five coconuts so it is impossible.
ReplyDeleteAfter reading the question many times, I figured out that the answer is 6. This is possible because the first person made 5 piles of 5, then he gave one to the monkey.
ReplyDeleteARE U KIDDING ME! I JUST DID IT AND MY COMPUTER DELETED WAT I SAID, RARGH, NYWAYZ so basically there are six coconuts each man gets 1 the monkey gets 1, that is abt all, now the real question is, WILL THEY SURVIVE, OR WILL THE MONKEY GET THEM ALL, AND REALLY WHO HAS A FIVE SEAT PLANE AND CRASHES ON AN ISLAND? I DONT EVEN THINK IT CULD FIT! yogi bear out
ReplyDeletein each pile the first time there is a minimum 6 coconuts in the very first pile.
ReplyDeleteYour friendly neighborhood Spiderman
The first pile has to be a multiple of 6 right? So for example 36, 36 divided by 5 = 7, with one left over. Then you subtract 8 coconuts(1 coconut for the monkey and one of the men takes 7) from 36, so you have 28 left. But that isn't a multiple of 6, so it doesn't work. What is also annoying is that the men do it 5 times, so it has to be a pretty big number. Using that info I put it into an excel spread sheet and I used an equation to figure it out. I've gone up to 566, and there is no equal number! Can there be such things as 1/8 of a coconut????
ReplyDeleteAccording to the information given in the text, the answer of 6 coconuts is impossible. This is because 6-1( the coconut given to the monkey) ,= 5 and then divided by 5 =1 and 5-1 =4. Then 4-1=3, but 3 cannot be divided by 5. If i were to try to figure out the answer, it would take me 5 or 6 hours to figure it out. I dont have the time, willingness or geniusness that enables me to spend that long on a math question and therefore i dont have an answer.If people think think that there is a trick to this, it is impossible because 6 only works for the original man.
ReplyDeleteguys, i'm going to help you out here... the question asks... what was the minimum the FIRST man could have had...
ReplyDeleteNow I get it... the question said that the smallest number the first man could of had. So yeah, the answer is six.
ReplyDeleteI figured out that the answer was six. The reason for this is that if the stranded guy puts the coconuts into piles and gives one to the monkey, there will be five left over. Since the question only stated, to quote, "In the original pile" it is obvious that six coconuts is the correct answer.
ReplyDeleteThe original number of cocunuts is: 12495. Man 1 took 2499 cocunuts, and the monkey took 1cocunut. This meant: 12495 – 2500(2499+1) = 9995 cocunuts. Man 2 took 1999 cocunuts, and the monkey took 1. This meant: 9995 – 2000(1999+1) = 7995 cocunuts. Man 3 took 1599 coconuts, and the monkey took 1. This meant: 7995 – 1600(1559+1) = 6395 cocunuts. Man 4 took 1279 coconuts and the monkey took 1. This meant: 6395 – 1280(1279+1) = 5115 cocunuts. Man 5 took 1023 coconuts and the monkey took 1. This left 5115 – 1024(1023+1) = 4091 cocunuts.
ReplyDeleteThis number is divisible by 5, because: 12495 divided by 5 = 2499. 5 for 1 big pile becoming 5 smaller piles, this shows that my total number is correct.
Adele
There would be 6 coconuts to begin with. At first I thought it would be almost impossible, but then I read the question carefully. Each man got 1 coconut, and 1 was given to a monkey.
ReplyDeleteok according to iris's theory i got my first answer worng, if it was only for the first man it would be six because, there would be 1 cocunut in each pile(5piles)(1+1+1+1+1) than the last cocunut would be for the monkey to take away!! yay!!!so in total that is 6 cocunuts.
ReplyDeleteadele
Ok, I know this is last minute, but I had to do some thinking. I can give two answers.
ReplyDeleteMy first answer is based on the orginal pile. The orginal pile meaning the last pile, or the first man. (As Iris said). He would have 6 coconuts, because there were 5 men so there would have to be at least 5 coconuts plus the monkey's coconut which equals a total of 6.
My second answer is based on how many coconuts there were to start with. I used an excel spread sheet and put in a formula (y = the number) (y - 1) -(0.2 x (y -1). Which I did to the first, second, third, fourth and fifth pile. After a few days of typing in multiples of six (starting at 6). I arrived at a number that actually worked. 3121!
Also, the first answer is theoretically possible and the second answer is mathematically possible. I don't know which one is the actual answer, so therefore I am putting both.
The lowest amount would be 6 because then 1 coconut in each pile.Then one for the MONKEY!!!!!!!seriously who in the world gives a good coconut to ................a.................MONKEY.........................
ReplyDelete$$$EM$$$
Let the original pile have n coconuts. Let a be the number of coconuts in each of the five piles made by the first man, b the number of coconuts in each of the five piles made by the second man, and so on.
ReplyDeleteWriting a Diophantine equation to represent the actions of each man, we have
n = 5a + 1 n + 4 = 5(a + 1)
4a = 5b + 1 4(a + 1) = 5(b + 1)
4b = 5c + 1 4(b + 1) = 5(c + 1)
4c = 5d + 1 4(c + 1) = 5(d + 1)
4d = 5e + 1 4(d + 1) = 5(e + 1)
Hence n + 4 = 5 × (5/4)4 (e + 1), and so n = (55/44) (e + 1) − 4.
Note that, since 5 and 4 are relatively prime, 55/44 = 3125/256 is a fraction in its lowest terms. Hence the only integer solutions of the above equation are where e + 1 is a multiple of 44, whereupon d + 1, c + 1, b + 1, and a + 1 are all integers.
So the general solution is n = 3125r − 4, where r is a positive integer, giving a smallest solution of 3121 coconuts in the original pile