Fun counting math puzzle

I just did this little math puzzle for an assignment in my discreet structures class and I thought I'd share it with you guys, let's see who can get the correct answer first.

The story goes like this: There are five pirates partying over a chest of diamonds they've just plundered which they're planning on splitting up in the morning before they go their separate ways. During the night one of the pirates wakes up and decides to go ahead and take his share so he can set sail early, so he splits the diamonds equally into five piles. At the end of the five piles he ends up with one extra diamond, so he thinks about it for awhile and decides to throw it in the ocean, take his pile, pool the rest of the diamonds back into the chest, and take off with his share.

A few minutes later, another pirate wakes up and does the exact same thing, not noticing that the first pirate had already taken his share, thus making five piles again. At the end, he too ends up with one extra diamond and decides to toss it into the ocean. After taking one of the piles and pooling the rest back into the chest, he sets sail as well.

This goes on one by one until all five pirates have eventually left, never noticing that the others left before them, and each time there is one extra diamond.

The question is: what is the least possible number of diamonds that could have possibly originally been in the chest? GOOD LUCK

oh, that's a nice puzzle.

Can these diamonds be broken into halves, or is the answer a whole number?

The answer is a whole number, assume the diamonds are all exactly the same size and value.

I'm assuming that the final pirate doesn't throw a diamond into the ocean.

YES THE FINAL PIRATE THROWS A DIAMOND!

Quote:

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Originally Posted by Necronopticous

YES THE FINAL PIRATE THROWS A DIAMOND!

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What an idiot.

he didn't know his friends left
honor amongst thieves!

The thing is that they all make five piles, and they all throw one extra diamond away because they don't realize the others already did the same thing.

Anyway, this is a math puzzle, it only follows that the characters involved would be completely stupid and not in any kind of sound mind.

This is stupid. I hate you, Jeremy.

I can't believe I worked for an entire page to try and figure this out. I couldn't even get past the second to last pirate.

I got up to four pirates when I ran into a number not divisible by four.

i'm still at 3

:choc:

AHAHAH!!! YAY! :kaolove2:

Yeah, I figured it out. It took a while to get the graph going and write so small but I pulled through.

(SPOILER)They started with 3,121 diamonds?

Quote:

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Originally Posted by Xaven

AHAHAH!!! YAY! :kaolove2:

Yeah, I figured it out. It took a while to get the graph going and write so small but I pulled through.

(SPOILER)They started with 3,121 diamonds?

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Thank god I stopped trying to figure it out by hand.

I should have never opened this thread. Damn you Necro! This thing is going to plague me now.

I'm not even gonna try it. It's a nice puzzle though, I just don't like doing math.

man Xaven
did you derive or did you graph the original, because deriving this would have a chain of like 1543443264326

:cry:

brb slitting writsts

OMG! Xaven is right. I had it when I said I had up to four pirates and just didn't realize it. (SPOILER)I just got to 2496 and said, "oh, that doesn't work because 2496/4 = 624 and I need a number that ends in 9" forgetting that all I had to do was multiply 624 by 5 then add one to get the answer rather than continuing on another step. I just worked it out by trial and error, recognizing some patterns so I really only had to try about 15 sets of numbers.

Anyone want my work?

Yes!

I made a grid like so:
(SPOILER)
3121 | 624
2496 | 499
1996 | 399
1596 | 319
1276 | 255


of course I got to top right and said "oh, this won't work" even though I was one step from the answer, because I am dumb.

And Del pats himself on the back for others' accomplishments.

Is that how it works in Staff, too? :p

Whoops. Just deleted my post. Del, can you put it back?

I'll add the pictures in a sec...

and I do Del's work for him!

Quote:

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Originally Posted by Raistlin

And Del pats himself on the back for others' accomplishments.

Is that how it works in Staff, too? :p

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Way to make me look bad Xaven.

But yes.

ahahahahahahahahahahaha

xDDDD

That deleted post was brilliantly timed.

Good job, Xaven. You shall now forever be known in my heart as a good man.

And here they are, big as ever.
http://i13.photobucket.com/albums/a2...ke/Workkk2.jpg
Tada? Yeah, not too impressive.

The other picture is of the other side of the paper which I did first. I didn't quite know what I was doing at the time hence the unposted scribbles.

so is there a more direct way to solve this other than trial and error? Some solution involving modular arithmetic?

Quote:

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Originally Posted by eestlinc

so is there a more direct way to solve this other than trial and error? Some solution involving modular arithmetic?

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Whatever that means.

I just looked for patterns. (SPOILER)Like how all the first steps increase by 20, and the seconds by 25. The fifth one is the one we needed so just use the almighty magic formula to work your way up the ladder 3 times (twice to find the difference and thrice to check). Find the difference 'tween the numbers playing the same role (48.828125 for the fifth ones). Then go 93.65625 (the smallest fifth step) + 48.828125 and spam your equals key 'till a whole number pops up. The End.

Edit: If your confused, reread as I added some missing words.

Yeah, that's what I want to know.

I don't know how well this would work but wouldn't you just do a x y problem? ((((((((((X-1)/5)-1)/5)-1)/5)-1)/5)-1)/5)=y

I have no clue if it would work as I just pulled it out of my ass but it kind of looks right :erm:

EDIT: Upon looking at what I typed it is bizull[img]/xxx.gif[/img][img]/xxx.gif[/img][img]/xxx.gif[/img][img]/xxx.gif[/img]z.

^ that is what i've been trying but i think i screwed up the equation :{


(SPOILER)(x-1)*5/4 - 1/5 - ((x-1)*5/4 -1/5)*5/4 -1/5 - (((x-1)*5/4 -1/5)*5/4 -1/5)*5/4 -1/5 - ((((x-1)*5/4 -1/5)*5/4 -1/5)*5/4 -1/5)*5/4 -1/5 - (((((x-1)*5/4 -1/5)*5/4 -1/5)*5/4 -1/5)*5/4 -1/5)*5/4 -1/5

I tried that, ed, but it seemed like more work than just plugging in values.

Plugging in numbers. You just type it into a calculator and tell it what x should be equal to and let it do the math for you.

That's what I tried at first but I had trouble making whole numbers out of it. Anyway, bedtime.

Quote:

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Originally Posted by edczxcvbnm

Plugging in numbers. You just type it into a calculator and tell it what x should be equal to and let it do the math for you.

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That works, but you won't get a whole number unless you start with (SPOILER)1,276. And I doubt there's anyway you would know to start there.

But what is x equal to? :p

EDIT: in reply to ed

I didn't have a calculator though.

yeah ok that equation doesn't work
nor does the derivative

there is so much blood on my wrists right now

I have no clue what X is equal to. I am too lazy to go get the calculator and make educated guesses...or better yet just write the forumal into a math program that would go through the numbers until there is no remainder.

I wrote a sloppy PHP bruteforce program to solve it:

http://demented3d.com/nissen/upload/prog02.jpg

Here is it working, showing the answer:

http://demented3d.com/nissen/upload/try2.php

Sweet

(SPOILER)After banging my head and getting tounge tied I came up with 3125

I find it ironic that despite not paying attention to the problem I still came rather close to the answer, getting 3206.

Either I'm bad at pen and paper math, or this is just weird.

[edit- I'm just bad at multiplying, that's all. my real answer (wronger than before) is now 3906xD]

The program makes it clear as day. That was in my head but it's hard to work out by hand without that nifty modulus function.

I'm sure by "fun" you mean "I'm going to rape your brain".

I never understood how people could enjoy maths, it must just be one of those things, like how some people enjoy hurting small woodland animals.

P.S. This topic is proof of why ninja are better, pirates are morons who throw away diamonds.

Quote:

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Originally Posted by Lindy

I'm sure by "fun" you mean "I'm going to rape your brain".

I never understood how people could enjoy maths, it must just be one of those things, like how some people enjoy hurting small woodland animals.

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You speak the truth. :up:

It isn't math. It is the fun of solving a problem. I also like to harm small woodland animals...:D

Quote:

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Originally Posted by Lindy

I'm sure by "fun" you mean "I'm going to rape your brain".

I never understood how people could enjoy maths, it must just be one of those things, like how some people enjoy hurting small woodland animals.

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But math is fun!

...I'm sad now. :kaocry:

I'm not quite sure what the problem is asking, nor am I sure how to get to that solution. Someone tone it down, kthx.

Quote:

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Originally Posted by Doomie

I'm not quite sure what the problem is asking, nor am I sure how to get to that solution. Someone tone it down, kthx.

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They're asking 'How many diamonds did the pirates start out with?'.

5 pirates find some treasure (a bunch of diamonds). Oblivious to eachother, they each divide the pile into five and take their share, BUT there's always one diamond left over (eg, 31 diamonds divided by five equals 6, with a remainder of 1). They toss the extra diamond overboard. Okay, so after the first pirate takes his share and tosses the extra diamond, the second pirate does the same, not knowing that the Pirate 1 already took his share! (Pirates 2-5 all get screwed) This happens until all 5 pirates have taken one fifth of the pile away, each time with one left over.

As for solving it, there's their big, fancy, BORING way, and my UN-BORING way. Your choice.

I didn't understand both. but I was using a calculator which gave me decimals, which may be why I didn't understand it. Thanks a bunch. ;]

This is also sadly brute-forcish:

Code:

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#!/usr/bin/ruby

def find_num(x, try, max)
    # Output intermediary steps if you want
    # puts "#{x},#{try},#{max}"
    return false if x % 5 != 1
    return true if try == max
    return x if find_num( (x-1) * 4/5, try+1, max)
end

1.upto(99999999999) do |x|
    if(ans = find_num(x, 1, 5))
        puts "Answer: #{ans}"
        break
    end
end

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Assuming this program works (and I have no idea if it does) you can figure out an arbitrary number of pirates greater than 1. The answer to 8 pirates is (SPOILER)390621. The answer to 9 pirates is (SPOILER)1953121. The answer to 10 pirates is (SPOILER)9765621. This program appears to take exponentially more time as you increase the number of pirates linearly, which means it's a sucky algorithm. 11 pirates would take many minutes to solve.

sorry to tell u guys this but the least possble # is 1 because theres more then 1 dimoned(i cant spell good) in the chest to start with so thats the least possble # i think?

So the other pirates divide up nothing?

Quote:

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Originally Posted by Treize

sorry to tell u guys this but the least possble # is 1 because theres more then 1 dimoned(i cant spell good) in the chest to start with so thats the least possble # i think?

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Nope.

It says they throw the extra diamond overboard. If there was 1 diamond and it was discarded by Pirate 1, then what would Pirates 2, 3, 4, 5 throw overboard? It says they all throw 1 extra diamond overboard.

I ended up with 3906 too.

1*5 +1 = 6
6 * 5 + 1 = 31
31 * 5 + 1 = 156
156 * 5 + 1 = 781
781 * 5 + 1 = 3906

3906 is the answer to the problem where the pirate divides the diamonds into 5 piles, takes 4 of the piles and leaves 1 pile behind.

Ahh, good catch.