Here is a good problem for you.

I would say that they were in posession of quite a few ugly ones, as they can't seem to get rid of them.


Aah! Homework.....how I used to hate it.

If Dan gives two animals to Jack and six animals to Jim, they'll all have the same number of animals.

Since we like to help with homework rather than doing it for you, why don't you tell us how you think you should start.

So, how many chickens does Dan have?

Solution is 11, 7, 21

Let's mark Jack's number of animals with x, Jim's with y and Dan's with z.

If Jack gives 6 pigs to Jim and gets 1 for it, he will loose 5 animals, where as Jim will receive 5. Then we conclude:

y+5=2(x-5)
y+5=2x-10
y=2x-15

The same way goes with the second trade, so:

x+13=3(z-13)
x+13=3z-39
x=3z-52

And then again with the last:

z+3=6(y-3)
z+3=6y-18
z=6y-21

So, as we concluded:

y=2x-15
x=3z-52
z=6y-21


We now mix. So it goes like this:

x=3z-52
x=3(6y-21)-52
x=18y-63-52
x=18(2x-15)-63-52
x=36x-270-63-52
385=35x
x=385/35
x=11

y=2x-15
y=2*11-15
y=22-15
y=7

z=6y-21
z=6*7-21
z=42-21
z=21

So Jack had 11, Jim 7 and Dan 21 animals.

Let's do the check.

y+5=2*(x-5)
7+5=2*(11-5)
12=2*6 which is obviously equal.

The same thing with the other two trades.