## Monkey business

Ten people land on a deserted island. There they find lots of coconuts and a monkey. During their first day they gather 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 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 slightly bloodied coconut. The monkey conks the second 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 possible number of coconuts in the pile, not counting the monkey’s?

December 13th, 2010 at 1:02 pm

I’ve cracked this little coconut. Its an odd number, probably greater than 9. If you want more…..

December 13th, 2010 at 1:20 pm

2519 (LCM minus the monkey’s)

2520/10=252

2520/9=280

2520/8=315

2520/7=360

2520/6=420

2520/5=504

2520/4=630

2520/3=840

2520/2=1260

December 13th, 2010 at 1:27 pm

DP’s right. I was just a little behind in solving it.

Must be a huge island to find so many coconuts.

First time commenting but I’ve lurked on this site a bunch.

December 13th, 2010 at 2:10 pm

19

if he was one short of ten equal piles that means he has one less than a multiple of ten. 19 is one less than twenty, a pile of twenty would have given each survivor 2 coconuts. 19 is only divisable by one and itself, therefore every survivor except the last one would have tried to obtain the final coconut from the monkey. since they all would die, the last survivor would be left with all 19 coconuts.

December 13th, 2010 at 2:17 pm

Hi DP. 2519 it is.

Let c be the smallest number of coconuts.

c must satisfy c+1 = 0 mod (10,9,8,7,6,5,4,3,2)

(i.e. c+1 = 0 for all of those mod bases).

This means that the smallest mod base that satisfies all of those is

LCM(10,9,8,7,6,5,4,3,2) = 2520. (LCM is Lowest Common Multiple)

So c+1 = 0 mod (2520), so c = 2519 mod (2520), so c = 2519.

December 13th, 2010 at 2:18 pm

Dp, you found where the piles would be equal at all times, no matter how many survivors were left. if your answer were so, there would be ten equal piles right from the start and no one would have died trying to get the final coconut.

December 13th, 2010 at 2:22 pm

okay, would you mind explaining why 19 is not correct since it satisfies everything in the question and is the smallest that can do so.

December 13th, 2010 at 2:49 pm

Hi Anthony. There was only one pile of 2519 coconuts. It couldn’t have been divided into 10 equal piles.

After adding the monkey’s coconut we would have 2520 coconuts. The first chap wanted 252. The next chap wanted 1/9 th of 2520 = 280 etc. See DP’s table for the rest.

No number smaller than 2520 is divisible by 10,9,8,7,6,5,4,3 and 2.

19 doesn’t work as 19+1 = 20 and isn’t divisible by 9,8,7,6 or 3.

December 13th, 2010 at 2:55 pm

Hi Sora. Welcome aboard. Better luck next time.

December 13th, 2010 at 3:11 pm

So much for politely giving others a chance……2519 + Monkeys coconut = 2520 as per Chris’s post no 5.

December 13th, 2010 at 3:32 pm

Hi Karl. DP had done it already (post 2) and with a (brief) explantion and a full verification.

December 13th, 2010 at 3:45 pm

I meant I worked it out with the same logic steps you had… ie starting at x+1 = 0 mod (10,9,8,7,6,5,4,3,2)

December 13th, 2010 at 3:59 pm

Hi Karl. I’m sorry. Inexplicably, I just assumed (unconsciously) that the because the problem had been solved so quickly and completely, that it had already passed it’s best before date. I’m really not sure why I jumped in so quickly, I usually wait a bit longer.

I thought that I’d just flesh out the details. Although I posted the problem, I hadn’t solved it myself. I often post problems that I haven’t solved, although I usually have an inkling about how to do it.

December 13th, 2010 at 11:57 pm

the problem never mentions it is looking for a number that is devisable by all those numbers at the same time. it says to find the smallest number that is not devisable by those numbers, and it says to exclude the monkey. if 2519 is the answer you are looking for, than you truly need to rethink your question. because your “required answer” is way off from your directions.

December 14th, 2010 at 1:38 am

Hi Anthony. Because there were so many coconuts, the people hadn’t initially realised that they were one coconut shy {ouch} of actually being able to divide the pile into 10 equal piles – they simply hadn’t counted them at that stage.

When the first man tried to divide the pile, he found he was 1 shy of divisibility by 10. (e.g. 19 fits with that, but so does 9). However, he gets killed and obviously didn’t obtain the monkey’s coconut).

Now there are 9 people left, so the pile needs to be divided into 9 piles. But once again it turns out that we are 1 coconut shy. 9 and 19 no longer work. 89 is the smallest number that satisfies the problem so far. The next round (8 people) takes us to 359 and the next (7 people) to 2519. The rest automatically satisfy the problem description.

Almost the entire problem definition was devoted to introducing the fact that the original number of coconuts+1 had to be divisible by 10,9,8,7,6,5,4,3 and 2.

December 14th, 2010 at 9:40 am

Anthony,

I originally agreed with your objection about 19 working, and spent a few minutes trying to work exactly why it wouldn’t work.

It took until I re-read the problem to figure it out; while you are correct in saying that 19 is prime, and hence cannot be divided by 9, the problem specifies that when you try to divide by 9 you end up one coconut shy (couldn’t resist! ) of the number you need. When you try to divide 19/9, of course you see that you actually have one coconut too many.

Therefore you need the first number which is one less than a multiple of 10, 9, 8, 7, 6, 5, 4, 3, and 2, which has been shown above to be 2519. Hope that explains a bit better

December 14th, 2010 at 2:02 pm

Thanks Adam, that definitely explained better than the others who just felt like again restateing their answers.

December 14th, 2010 at 7:00 pm

Hi Anthony. Do you now accept that the question is very clear and doesn’t have the several faults that you deemed it to have?

I don’t know why you think that I merely restated my previous answer. I walked you through the problem and its solution, describing everything that was happening as I proceeded.

The only thing I didn’t do was to explain how I determined the relevant LCMs.

You still haven’t responded to my “challenge” in the “black or white” problem (post 47).

December 16th, 2010 at 4:06 am

smallest possible no of coconuts is 89 . . .

December 16th, 2010 at 4:11 am

sorry i was wrong… the smallest possible no would be 2519. . .

December 16th, 2010 at 5:11 pm

I say conk the monkey over the head and eat him as well.

December 19th, 2010 at 5:13 pm

reallly? so guy wakes up to find 5 dead guys, is able to do the math to see he has 503 coconuts, but for some reason he still wants to battle the war veteran bloddied monkey for an even number?

the real puzzle is a logical one!

December 19th, 2010 at 5:19 pm

its 99 fo suro

December 19th, 2010 at 9:37 pm

the answer to the question of what the lowest numbe of coconuts in the pile is stupidly simple

the answer being 1 because the monkey has the one required for 10 equal piles but every castaway took a coconut before getting conked on thehead and dyying so the number of coconuts in the original pile comes down to a lowest number of 1

XD

December 20th, 2010 at 1:24 pm

x=total pile; x-1=including monkey

smallest # in a pile would be x/10 or (x-1)/10.

Let the monkey have the coconut!

December 21st, 2010 at 2:39 pm

Here’s the math behind it:

So we have to figure out the multiplication of prime numbers of 10 through 2 (not 1 because anything is divisible by 1)

10 is 2*5

9 is 3*3 or 3^2

8 is 2*(2^2)

7 is 7

6 is 2*3

5 is 5

4 is 2*2 or 2^2

3 is 3

2 is 2

Take the unique numbers that has the highest square and muliply each..

(2^2)*5*(3^2)*7

thus..

4*5*9*7 = 1260 (which incl the monkey’s coconut)

Answer is 1260-1 = 1259 coconuts in the pile.

-Paul

December 21st, 2010 at 3:52 pm

Hi Paul. 8 is 2^3.

Then to get the LCM, we need the highest powered prime of any (initial) factor. There is nothing special about the squares (for LCM purposes).

So LCM = 2^3 * 3^2 * 5 * 7 = 2520.

December 22nd, 2010 at 7:12 am

DOH! That’s what I get for not double-checking calculating LCMs. Yes, I should have done 2^3, not dropping off the 2. 1260 is not evenly divisible by 8 anyhow.

Good exercise. 2520 is the LCM of 10 to 2, thus 2519 in the pile.

Paul