
#1
Aug2607, 04:40 PM

P: 12

A new one for you guys: a math one and a bit different version of the Gilligan's island one:
Ten shipwrecked people land on a deserted island. There they find heaps of coconuts and a single monkey. During their first day they gather the coconuts and put them all in a community pile. After working all day they decide to sleep and divide them into ten equal piles the next morning. That night one castaway wakes up hungry and decides to take his equal share early. After dividing up the coconuts he finds he is one coconut short of ten equal piles. He also notices the monkey holding one more coconut. So he tries to take the monkey's coconut to have a total evenly divisible by 10. However when he tries to take it the monkey conks him on the head with it and kills him. Later another castaway wakes up hungry and decides to take his share early. On the way to the coconuts he finds the body of the first castaway, which pleases him because he will now be entitled to 1/9 of the total pile. After dividing them up into nine piles he is again one coconut short and tries to take the monkey's coconut. Again, the monkey conks the man on the head and kills him. One by one each of the remaining castaways goes through the same process, until the 10th person to wake up gets the entire pile for himself. What is the smallest number of possible coconuts in the pile, not counting the monkey's? have fun! 



#2
Aug2607, 07:31 PM

P: 880

funny (bad) monkeyhe's a NUTTER problem !!!
and that (problem) took longer than I exected 



#3
Aug2607, 07:55 PM

P: 1,520

That's a no brainer...
The product of the greatest powers inferior to ten of the primes inferior to 10 (5*7*8*9) minus one. 



#4
Aug2607, 08:40 PM

P: 880

.:Deserted island:.If this was a question on a test, and that was your explanation on/for your answer, you probably wouldn't score very well. (eight is a cubed number)  and actually there is another mistake in your answer, too 



#5
Aug2707, 12:22 AM

P: 1,520

Well, I'd like a second opinion on that, it's perfectly clear and correct to my eyes.




#6
Aug2707, 08:26 AM

P: 880

let's just say, the answer may be correct, but that isn't how you solve the problem




#7
Aug2707, 11:19 AM

P: 1,520

If you say so. I'm satisfied with my answer, but I can't force you to be.




#8
Aug2707, 12:06 PM

P: 12

Werg, can you explain your answer a bit more? Sadly I dont get it.




#9
Aug2707, 12:07 PM

P: 880

It's from the number of letters in the words:  oh, yeah, and add one to the last word for the monkey's coconut 



#10
Aug2707, 12:23 PM

P: 403

Then, (N+1) = (2**3) * (3**2) * 5 * 7 = 2520 So, N=2519 . 



#11
Aug2707, 01:08 PM

P: 1,520

Really, rewebster, why do you want me to make this uselessly long? You know there's only one way to solve this and you know I solved it the right way. I didn't use magic to get the answer.




#12
Oct1507, 01:00 AM

P: 6

The number should be something which when divided by 2..10 gives remainder 1,2,3..9, right?
That's LCM of (2,..10) 1 



#13
Oct1607, 12:06 PM

P: 640





#14
Oct1607, 07:05 PM

P: 106

No, he said 8 is one of the greatest powers inferior to ten of the primes inferior to 10.
And 8 is, de facto, the greatest power of 2, inferior to 10. It is OK. 



#15
Oct1707, 10:45 AM

P: 657

1) Find all primes less than 10: 2, 3, 5, 7. 2) Find the greatest powers of those primes such that the result is less than ten: 2^3 = 8, 3^2 = 9, 5^1 = 5, 7^1 = 7. Hence, 8,9,5,7. 3) Find the product of 8*9*5*7. That's the LCM of all the numbers 110. 4) Subtract 1 (the monkey's). DaveE 


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