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Apples and Monkeys


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31 replies to this topic

#11 heatherlovesjade

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Posted 03 March 2008 - 11:57 AM

10- fiive monkeys all wake up eat one and hide one- 5 times number of monkeys= 5 +5 hidden = 10
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#12 mohamdy

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Posted 04 March 2008 - 03:34 PM

Sorry
i Asked the owner and said 1 extra

If so 3121 is the smallest number
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#13 Duh Puck

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Posted 04 March 2008 - 03:49 PM

Sorry
i Asked the owner and said 1 extra

If so 3121 is the smallest number

Good try though. I don't know if English is your first language, but if not, then that misunderstanding was very ... understandable. The difference between your interpretation and the poster's intent was due only to the word "an" before "extra". If you remove that one word, your solution would be correct. ;)
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#14 infophiliac

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Posted 04 March 2008 - 05:51 PM

After a little math, you can show that the final share has to be 1 less than a power of 2.
A simple program to calculate the other shares does not give integral values for 15, 31, 63 or 127,
but does for 255: -> 255 319 399 499 624 ---> 3121.


I'm a bit confused about the "little math" and "simple program" implicitly needed to solve this problem. :wacko:

Let me explain what I had figured out on my own, and then perhaps someone can tell me where I went wrong:
  • Let A0 equal the number of apples at start, with A1-5 being the number of apples remaining after each monkey takes its share.
  • Let S1-5 equal each monkey's respective share.
  • Let x equal the number of monkeys whom have so far interacted with the apples.
Because there is an extra apple when each monkey divides the total quantity by 5, but given that we are looking for a minimum starting quantity, it follows that A0-4 would have to be one greater than a multiple of 5, while A5 would merely be a multiple of 5. Easy enough.

So where (x < 5)...
  • Ax+1 = (Ax - 1) - Sx+1
  • Sx+1 = (Ax - 1) / 5
And, with any value of x...
  • Ax+1 = Sx+1 * 4
It is with this final equation that I began looking for the solution:
In order to get prospective Sx values, I started looking for numbers that when multiplied by 4 would yield a product one greater than a multiple of 5. When I found some, I started looking for a rule, and, lo and behold, I found one: the product of 4 and any number ending in 4 or 9 fits the desired criteria. So I gruelingly started plugging numbers ending in 4 or 9 into the final equation above, then plugged the result into the second equation above hoping that it would yield a result that also fit the criteria. If it did, I could plug it into the final equation, then into the the second equation, and, insofar as the criteria kept being met, so on and so forth for 4 consecutive iterations before I hit a number that was a straight multiple of 5 (A5). If I found such a number (S1) then I would merely have to multiply it by 5 and add 1 in order to get the solution (A0).

The number I was apparently looking for was 624: 624 * 5 + 1 = 3121. Needless to say, I didn't make it through the 65 trials I would have had to go through by my method in order to get there. Granted, I do know enough programming that I could have written a little script that would have come up with the solution for me. Alas, I figured that would be cheating. Further, while my grueling trial-and-error method would have worked, I converged not at all on the realization that the final share would have to be one less than a power of 2, nor did I have the knowledge requisite in order to seek out so-called integral values. So I'm left curious as to what exactly I was missing.

Now, while I don't know any calculus myself, I know enough about calculus to be able to tell what kind of problems require it, and this one seemed to fit that criteria. However, I convinced myself that obtaining the solution was probably--as it so often is with puzzles on this site--much easier than I was making it out to be. With the use of terms like "integral value" in the quoted explanation above, though, it seems I stand corrected.

So, while I don't expect a calculus lesson from anyone, if someone could please tell me generally what I needed to know in order to solve this problem efficiently, it would be very much appreciated!

Thank you! :D
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#15 keisha4

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Posted 11 March 2008 - 12:24 AM

tegrfdbgnhj;oiuytrfdxc nm,'[p0o9iuyr4e3w2q N

Edited by keisha4, 11 March 2008 - 12:25 AM.

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#16 Lost in space

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Posted 11 March 2008 - 10:38 AM

There is no way the monkeys can calculate. They can however write the complete workds of Shakespear if you have an infinite number of monkeys and typewriters. asld;gkabvowegwegw'vj
To be or not to be? That is the gasdlhkfsdf.
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#17 doom48

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Posted 14 March 2008 - 05:53 PM

did you guys remember the 1 monkey took his share and hid it while the other monkeys were sleeping how did they know he took his shsre?
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#18 babbal

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Posted 15 March 2008 - 09:10 PM

did you guys remember the 1 monkey took his share and hid it while the other monkeys were sleeping how did they know he took his shsre?

they don't. They think that they're still the first.
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#19 Kapil

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Posted 19 March 2008 - 05:48 PM

I believe the correct answer should be 3281

1st monkey finds 3281 apples, devides in group of 656 each and eats 1.
2nd monkey finds 656 apples, devides in group of 131 each and eats 1.
3rd monkey finds 131 apples, devides in group of 26 each and eats 1.
4th monkey finds 26 apples, devides in group of 5 each and eats 1.
5th monkey finds 5 apples, devides in group of 1 each and eats 1.
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#20 braindraino

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Posted 13 April 2008 - 08:33 AM

Yep - 3121!
monkey 1 finds 3121, eats 1, gives each 624, hides his share
monkey 2 finds 2496, eats 1, gives each 499, hides his share
monkey 3 finds 1996, eats 1, gives each 399, hides his share
monkey 4 finds 1596, eats 1, gives each 319, hides his share
monkey 5 finds 1276, eats 1, gives each 255
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