Deserted Island Math Challenge: Solve for the Coconuts

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The discussion revolves around a mathematical puzzle involving ten shipwrecked people and a monkey on a deserted island, where they attempt to divide coconuts. Each castaway, upon waking, finds the total number of coconuts insufficient for equal division, leading to their demise when they try to take the monkey's coconut. The participants debate the correct method to determine the smallest number of coconuts, with one user asserting that the answer is derived from the least common multiple of numbers 2 through 10, minus one. Ultimately, the conclusion reached is that the minimum number of coconuts is 2519. The conversation highlights the importance of logical reasoning and clarity in mathematical problem-solving.
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.:Deserted island:.

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!
 
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funny (bad) monkey---he's a NUTTER problem !---



---and that (problem) took longer than I exected
 
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.
 
Werg22 said:
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.


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
 
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Well, I'd like a second opinion on that, it's perfectly clear and correct to my eyes.
 
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let's just say, the answer may be correct, but that isn't how you solve the problem
 
If you say so. I'm satisfied with my answer, but I can't force you to be.
 
Werg, can you explain your answer a bit more? Sadly I don't get it.
 
Werg22 said:
If you say so. I'm satisfied with my answer, but I can't force you to be.

OK, but getting an answer from the problem itself, it would be about the same as saying:

It's from the number of letters in the words:

Hanae4fun said:
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!

In other words, there seems to be no logic or a proof from what that formula is derived or how it is formed from the problem to get the answer---if that doesn't bother you--then the answer given by the bold letters is an accepted proof too
----

oh, yeah, and add one to the last word for the monkey's coconut
 
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  • #10
Hanae4fun said:
What is the smallest number of possible coconuts in the pile, not counting the monkey's?

Well, the number of coconuts plus one (N+1) is the minimum common multiple of 2,3,4,5,6,7,8,9 and 10.
Then, (N+1) = (2**3) * (3**2) * 5 * 7 = 2520
So, N=2519 .
:smile:
 
  • #11
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
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
Werg22 said:
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.

Are you saying that 8 is a prime?
 
  • #14
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
K.J.Healey said:
Werg22 said:
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.
Are you saying that 8 is a prime?

Just to clarify:

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 1-10.
4) Subtract 1 (the monkey's).
[/color]

DaveE
 
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