Split the booty

[Guest]

A pirate crew at the end of the day split the booty. The first pirate got 100 gold pieces, and 1/6 of the remaining booty. The second one got 200 gold pieces, and 1/6 of the remaining booty. The third one got 300 gold pieces, and 1/6 of the remaining booty. Ect. The last one only got, what if left from the booty.

At the end, every pirate had the same ammount of gold pieces (from the booty).

How many pirates were there, and how much was the booty.

[Guest]

If I solved it right, then:

Gold: 2500

Pirates:5

[bonanova]

I worked a long puzzle earlier today and didn't have the heart for this one, although it's less complicated than the one I did do.

Aren't you glad I shared that?

Very cute, and yes, I'd say you solved it correctly.

[Guest]

I'd say the answer is a total of 3000 gold split between 5 pirates.

Solved:

100 + (1/6)*total_gold = Pirate_1

200 + (1/6)*(total_gold - {Pirate_1}) = Pirate_2

Therefore:

200 + (1/6)*(total_gold - 100 - (1/6)*total_gold) = Pirate_2

200 + (1/6)*((5/6)*total_gold - 100) = Pirate_2

200 + (5/36)*total_gold - 16.667 = Pirate_2

183.333 + (5/36)*total_gold = Pirate_2

Now, everyone ends up with equal amounts, so:

Pirate_1 = Pirate_2

100 + (1/6)*total_gold = 183.333 + (5/36)*total_gold

(1/36)*total_gold = 83.333

total_gold = 3000

If the total is 3000, then

Pirate_1 = 100 + 3000(1/6)

Pirate_1 = 600

The total number of pirates must therefore be:

(3000 total gold) / (600 per person) = 5 Pirates

[Guest]

5 pirates, 2500 gold

1st pirate gets 100+1/6(2400)=500 gold

2nd pirate gets 200+1/6(1800)=500 gold

3rd pirate gets 300+1/6(1200)=500 gold

4th pirate gets 400+1/6(600)=500 gold

5th pirate gets 500 gold

[Guest]

I see my error. My math does not account for the last pirate not getting a set amount. My solution would only work if the last pirate were given a set amount of 100 more than the previous plus 1/6 the total gold.

[Guest]

2500 gold

5 pirates

Set the initial 2 pirates gold equal as follows:

x = total gold

100 + [(x-100)/6] = 200 + [(x - 200 - 100 - ((x-100)/6)/6]

Rearranging and solving:

x - 100 = 300 + x - [(x-100)/6]

And again:

0 = 400 - [(x-100)/6]

And again:

0 = 2400 - x + 100

Therefore x = 2500

Substitue this value into the equation for Pirate 1

[Guest]

A pirate crew at the end of the day split the booty. The first pirate got 100 gold pieces, and 1/6 of the remaining booty. The second one got 200 gold pieces, and 1/6 of the remaining booty. The third one got 300 gold pieces, and 1/6 of the remaining booty. Ect. The last one only got, what if left from the booty.

At the end, every pirate had the same ammount of gold pieces (from the booty).

How many pirates were there, and how much was the booty.

gold was 2500 pieces and there were 5 pirates. It really wasn't that hard.

[Guest]

this puzzle did a fantastic job at proving how horrible my algebra skills have become. I knew the formulas, I knew what I needed to do with the formulas, but it took six tries to get it right. thanks Vobnalb9. I'm gonna go sit in on some 9th grade classes now.

Edited March 19, 2008 by ALFRED

[Guest]

A pirate crew at the end of the day split the booty. The first pirate got 100 gold pieces, and 1/6 of the remaining booty. The second one got 200 gold pieces, and 1/6 of the remaining booty. The third one got 300 gold pieces, and 1/6 of the remaining booty. Ect. The last one only got, what if left from the booty.

At the end, every pirate had the same ammount of gold pieces (from the booty).

How many pirates were there, and how much was the booty.

Every pirate gets part of the remaining booty, not a part of all the booty, therefore (x being the starting booty):

100 coins and 1/6 of remaining booty (x-100)

First pirate gets 100 + 1/6 (x-100)

200 coins and 1/6 of the remaining booty (x-a total of what the first pirate got (100 + 1/6(x-100) - 200 coins)

second pirate gets 200 + 1/6 (x-200-100-1/6(x-100))

First pirate got as much as the second pirate:

100+1/6(x-100) = 200 + 1/6(x-200-100-1/6(x-100)) |open parenthesis

100+1/6x-100/6 = 200 + 1/6(x-200-100-1/6x+100/6) |open parenthesis

100+1/6x-100/6 = 200 + 1/6x - 200/6 - 100/6-1/36x+100/36 |1/6x and -100/6 delete from both sides of the equasion

100 = 200 - 200/6 - 1/36x + 100/36 |x to the right, numbers to the left

1/36x = 100 - 200/6 + 100/36 |multiply with 36

x = 3600 - 1200 + 100

x = 2500

The remaining part of the riddle was not necessary to solve equasion with one unknown only one equasion is needed;

Edited April 24, 2008 by comcast

[Guest]

A pirate crew at the end of the day split the booty. The first pirate got 100 gold pieces, and 1/6 of the remaining booty. The second one got 200 gold pieces, and 1/6 of the remaining booty. The third one got 300 gold pieces, and 1/6 of the remaining booty. Ect. The last one only got, what if left from the booty.

At the end, every pirate had the same ammount of gold pieces (from the booty).

How many pirates were there, and how much was the booty.

[Guest]

While I'll not argue the mathematical accuracy of the answers provided, I do want to throw in a real world caveat. The riddle refers to the pirates as the First, second, third, and so on. The answers given only work if we divide the total loot mathematically before paying all 5 pirates. But then this would not be a likely scenario given that each pirate is receiving a different amount up front, and then a standard 1/6th share of remaining loot.

If the riddle implies a sequential pattern to payment (first, second, ect) then the solution of 2500 gold per 5 pirates, or a total before splitting of 12500 will not result in each pirate receiving the same amount. Think of it like balancing a checking account.

12500 - 100 = 12400 remaining

1/6th of 12400 is 2066.6666666667. So the first pirate would receive 100 + 2066.66666667 or 2166.6666667.

The remaining funds in the loot would be 12500-2166.66666667 or 10333.3333333

Then the second pirate is paid.

10333.3333 - 200 = 10133.3333333 remaining

1/6th of 10133.33333 is 1688.8888888889. So the second pirate would receive 200 + 1688.888888889 or 1888.888888889

And so on....

If someone's up for the challenge, I'd enjoy seeing a more elegant mathematical expression that solves this riddle within the literal context.

[Guest]

Yarrrr, but the solution isn't that each of the 5 pirates receives 2,500, but that each of the 5 pirates receives 1/5 x 2500 = 500g

Pirate 1:

(a) Starting total: 2500

(b) Upfront gold: 100

( c) a-b: 2400

(d) 1/6 x c = 400

(e) b + d = 500

Pirate 2:

(a) Starting total: 2000

(b) Upfront gold: 200

( c) a-b: 1800

(d) 1/6 x c = 300

(e) b + d = 500

Pirate 3:

(a) Starting total: 1500

(b) Upfront gold: 300

( c) a-b: 1200

(d) 1/6 x c = 200

(e) b + d = 500

Pirate 4:

(a) Starting total: 1000

(b) Upfront gold: 400

( c) a-b: 600

(d) 1/6 x c = 100

(e) b + d = 500

Pirate 5:

(a) Starting total: 500

(b) Upfront gold: 500

( c) a-b: 00

(d) 1/6 x c = 00

(e) b + d = 500

but I couldn't type it, it's rated arrrrrrr

[Guest]

Ahh. Just goes to show that I wasn't reading the responses in the right context. Thanks for setting me straight.

[Guest]

A pirate crew at the end of the day split the booty. The first pirate got 100 gold pieces, and 1/6 of the remaining booty. The second one got 200 gold pieces, and 1/6 of the remaining booty. The third one got 300 gold pieces, and 1/6 of the remaining booty. Ect. The last one only got, what if left from the booty.

At the end, every pirate had the same ammount of gold pieces (from the booty).

How many pirates were there, and how much was the booty.

I believe it can be solved by the formula: 100+1/6x = 200 +5/36X

X=3600 and 6 pirates