Apples and Monkeys

[Guest]

There are 5 Monkeys and a number of apples. They want to divide the apples into equal shares, but all forget. They all go to sleep. In the middle of the night, the first monkey suddenly remembers, and goes and divides the apples into five equal groups. He finds out that there is an extra, and eats it. Then, he hides his share and goes to sleep. The second monkey remembers ten minutes later, divides the apples, and finds out there is an extra. He eats the extra, hides his share and goes to sleep. The third, fourth, and fifth all do the same process, and at the end, there are still many apples left. Assuming that the monkeys didn't cut any apples, what is the least number of apples that there could've been at the beginning?

[Guest]

3121. Monkey 1 sees them and divides them into 5 shares of 624. He hides his share and eats the left over. Monkey 2 sees the 4 shares of 624 and divides them into 5 shares of 499. He eats 1 and hides his 499. Monkey 3 sees the 4 shares of 499. He divides them into 5 shares of 399 and eats 1. He hides his 399. Monkey 4 sees the 4 shares of 399. He divides them into 5 shares of 319 and eats 1. He hides his 319. Monkey 5 sees the 4 shares of 319. He divides them into 5 shares of 255. He eats the left over and hides his 255. There are 1020 left over.

[Guest]

duuude.

That's a lot of apples

[Guest]

Anyone find a lower one that divides into whole apples?

[bonanova]

Anyone find a lower one that divides into whole apples?

No. Your answer is the right one.

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.

[Guest]

how could you forget about 3121 apples laying in front of your face i know their monkeys but please 3121 APPLES !! how could you !!!!

[Guest]

The Number is 31 apples

MONKEY 1 sees 31 apples he eats 1 and distribute 30 into 5 groups of 6 apples hides his 6 apples and leave 24 apples

MONKEY 2 sees 24 apples he eats 4 and distribute 20 into 5 groups of 4 apples hides his 4 apples and leave 16 apples

MONKEY 3 sees 16 apples he eats 1 and distribute 15 into 5 groups of 3 apples hides his 3 apples and leave 12 apples

MONKEY 2 sees 12 apples he eats 2 and distribute 10 into 5 groups of 2 apples hides his 2 apples and leave 10 apples

MONKEY 1 sees 8 apples he eats 3 and distribute 5 into 5 groups pf 1 apple hides his 1 apple and leave 4 apples

I first thought of 26 but mokey 2 would have 20 apples which will be distributed and he would not have any apple to eat

Edited February 28, 2008 by mohamdy

[Guest]

For the second monkey you have taken away 4 then taken another 4 away so atotal of 8 overall;

He ate 4 then hid 4.

Thats one gready monkey .

MONKEY 2 sees 24 apples he eats 4 and distribute 20 into 5 groups of 4 apples hides his 4 apples and leave 16 apples

[Guest]

The Number is 31 apples

MONKEY 1 sees 31 apples he eats 1 and distribute 30 into 5 groups of 6 apples hides his 6 apples and leave 24 apples

MONKEY 2 sees 24 apples he eats 4 and distribute 20 into 5 groups of 4 apples hides his 4 apples and leave 16 apples

MONKEY 3 sees 16 apples he eats 1 and distribute 15 into 5 groups of 3 apples hides his 3 apples and leave 12 apples

MONKEY 2 sees 12 apples he eats 2 and distribute 10 into 5 groups of 2 apples hides his 2 apples and leave 10 apples

MONKEY 1 sees 8 apples he eats 3 and distribute 5 into 5 groups pf 1 apple hides his 1 apple and leave 4 apples

I first thought of 26 but mokey 2 would have 20 apples which will be distributed and he would not have any apple to eat

Each monkey ate 1 apple while dividing...you are wrong!!

[Guest]

In the puzzle

He Says He finds out that there is an extra, and eats it

he did not say there is 1 extra .

i think here that "there is an extra" means that the extra number is unknown

and "eats it" means he eats this extra

[Guest]

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

[Guest]

Sorry

i Asked the owner and said 1 extra

If so 3121 is the smallest number

[Guest]

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.

[Guest]

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.

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 A_1-5 being the number of apples remaining after each monkey takes its share.
  
• Let S_1-5 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 A_0-4 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)...
  
  • A_x+1 = (A_x - 1) - S_x+1
    
  • S_x+1 = (A_x - 1) / 5
    

  And, with any value of x...

  [*]A_x+1 = S_x+1 * 4

  It is with this final equation that I began looking for the solution:

  In order to get prospective S_x 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!

[Guest]

tegrfdbgnhj;oiuytrfdxc nm,'[p0o9iuyr4e3w2q N

Edited March 10, 2008 by keisha4

[Guest]

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.

[Guest]

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?

[Guest]

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.

[Guest]

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.

[Guest]

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

[Guest]

Yup, 3121 is what I found too through Excel.

Wow, if I need to use Excel to calculate the share, and it takes some minutes to figure out; how could the 5 monkeys can do counting (without mistakes, do they do double checking), dividing (again, gurantee no mistake) and hiding their individual share one by one in 1 overnight time?

Are they supermonkey?

Just kidding.

[Guest]

In the puzzle

He Says He finds out that there is an extra, and eats it

he did not say there is 1 extra .

i think here that "there is an extra" means that the extra number is unknown

and "eats it" means he eats this extra

He finds out that there is an extra, and eats it

an & it are singular therefore only 1

Edited May 19, 2008 by Mistwalker

[Guest]

In the puzzle

He Says He finds out that there is an extra, and eats it

he did not say there is 1 extra .

i think here that "there is an extra" means that the extra number is unknown

and "eats it" means he eats this extra

"an" is grammar for "a". the n is added to the "a" because of the following vowel "e" in extra. therefore a is equal to one. also, the word extra should be "extras" when speaking of the second monkeys findings. this riddle is either written incorrect or your answer is wrong.

[Guest]

In the puzzle

He Says He finds out that there is an extra, and eats it

he did not say there is 1 extra .

i think here that "there is an extra" means that the extra number is unknown

and "eats it" means he eats this extra

Solutions are of the form X + N*(5**5) . (N more apples for each at the end). X = -4 works. Start with -4 apples, the monkey first eats an apple then divides the -5 apples, takes -1 and leaves -4 for the next monkey. -4 is not a feasible solution, so add 3125 (5**5) for the same answer that Senior Member posted: 3121.

[Guest]

Hmm. I worked it out by pencil and paper but got 12496.

12496 works but it's not the smallest number. I made some subtle mistake about the form the starting and ending piles must be in.

[Guest]

Here's how I got my answer:

Let N(0) be the number of apples the FIRST monkey sees when he wakes up, N(1) is the number the second monkey sees, etc etc, and N(5) is the number that is left at the end of the whole slumber party.

All piles except the first and last pile must have a number N() of apples of the form

N(i) = p(i+1) * 5 + 1 = p(i) * 4

for some set of numbers p(i), where i ranges from 1 to 5. You can think of p(i) as the size of the sub-piles that monkey (i) makes when he sorts his apples.

The final pile is simply of the form

N(5) = p(5) * 4

and the first pile is of the form

N(0) = p(1) * 5 + 1

You can then verify that, for all p(i) to be integers, we must have for i ranging from 1 to 4 that

N(i) =(5 n(i) - 1) * 4 = (4 n(i) -1) * 5 +1 = 20 * n(i) - 4

for some integers n(i), where

p(i) = 5 * n(i) – 1 = 4 * n(i-1) -1.

Working backwards from a pile of such form, say N(4), it is necessary and sufficient that the previous pile correspond to n being increased by some integer multiple of 4, in order to ensure that the previous pile will remain in the stated form.

Call this quantity m(i) = n(i) – n(i+1). To restate, all m(i) except m(1) must be divisible by 4, and i runs from 1 to 3. (This is the subtle part that I missed: m(1) does not have to be divisible by 4. If you make m(1) divisible by 4, you get my answer of 12496.)

Each time we iterate backward, m also increases. (That is, m(2) > m(3), etc)

You can check that, each iteration going backwards, m must increase by a factor of 5/4. Since this must happen twice going from m(3) backwards to m(1), then m(3) must be divisible by 4^2, i.e, 16.

The smallest nonzero integer divisible by 16 is 16, so m(3)=16. I think this is both necessary and sufficient, but I haven't quite nailed the proof.

Anyway m(3) = 16 means that n(4) = 64, since you can verify that n(i) = 4 * m(i-1).

From that you can back out that N(0) = 3121. No computer needed if you don't make subtle logic mistakes.

I'm going home. My brain hurts. I'm sure I've made a couple typos in here since the notation I actually used on my notepad was COMPLETELY different and I probably messed the translation up somewhere.