The Two Thieves

Question

Fatty and Skinny are two thieves, who stole gold coins and now divide them between themselves. Skinny says, “That’s unfair, Fatty, in your bag there are three times more coins than in mine.” Fatty answers, “OK Skinny, I am giving you one more coin for each year of your age.” Skinny is still not happy, “But you still have double my coins.” “This is only fair, as I am twice your age,” Fatty concludes. Later, Fatty leaves for a short time, and Skinny takes one coin from Fatty’s bag for each year of Fatty’s age, and puts these coins in his own bag. Now, which bag has more coins, Fatty’s or Skinny’s?

Answer 1

The answer is:

Now they each have an equal number of coins.

At the start

Let's say Skinny is y years old. So Fatty is double that: 2y.

Let's say Skinny has x coins at the start. Fatty has triple that: 3x.

Fatty then gives Skinny y coins, so Skinny has x+y coins, and Fatty has 3x-y. We are told Fatty still has double.
So, 2(x+y) = 3x-y ... x=3y

Then Skinny takes 2y coins. Now Skinny has x+y+2y = x+3y = 6y, and Fatty has 3x-y-2y = 9y-3y = 6y.

Answer 2

We can represent the relationships with three variables:

f - the starting coins for fatty
s - the starting coins for skinny
a - the age of skinny (the age of fatty is 2a)

The formula at the start:

Fatty has 3 times the coins of Skinny

f = 3s

After fatty gives skinny a number of coins equaling skinny's age:

Fatty (f) minus skinny's age (a) is still twice Skinny (s) + skinny's age (a)

f - a = 2(s + a)

In terms of Fatty's starting coins:

Simplify f - a = 2(s + a)

f - a = 2s + 2a
f = 2s + 3a

in terms of s since f is 3s

3s = 2s + 3a
s = 3a

So

s = 3a and f = 3s so at at the start Skinny has 3a coins and Fatty has 9a coins.
After Fatty giving away a coins, Fatty has 8a and Skinny with 4a because Fatty then has double the coins.

After Skinny takes another 2a coins, Fatty has 6a and Skinny has 6a coins.

Therefore after all is said and done, they have the same number of coins.