+870 "If you want to stay alive,
Gimme half of your money,
Plus twenty more pounds.
Or else..."
The unfortunate merchant obeys. Then the thief runs away.
A few minutes later, the merchant meets another thief, which asks him the same thing as the first one.
He meets a third one, and a fourth one, and finally a fifth one.
When he finally reaches his house, his coin purse is empty. (Poor merchant 
)
How many coins did he have at the beginning of his travel ?
You will be marked out of 20, and get a cookie if your mark is 16 or more 
+128460 Let's call x, the total amount he started with before his unfortunate fate {I'm assuming that a coin = a pound}
The first thief gets (1/2)x + 20.......and he is left with x - [(1/2)x + 20)] = (1/2)x - 20
The next thief gets (1/2)[ (1/2)x - 20] + 20 = (1/4)x + 10 .....and he is left with :
x - [ (1/2x + 20] - [(1/4)x + 10 ] = (1/4)x -30
The third thief gets (1/2)[(1/4)x - 30)] + 20 = (1/8)x + 5 and he is left with
x - [(1/2)x + 20] - [(1/4)x + 10] - [(1/8)x + 5] = (1/8)x - 35
The fourth thief gets (1/2)[(1/8)x - 35] + 20 = (1/16)x + 2.5 and he is left with
x - [(1/2)x + 20] - [(1/4)x + 10] - [(1/8)x + 5] - [(1/16)x + 2.5] = (1/16)x - 37.5
The last thief gets (1/2)[(1/16)x - 37.5] + 20 = (1/32)x + 1.25
And he is left with....uh......nothing !!!!
So we need to solve the following equation :
x - [(1/2)x + 20] - [(1/4)x + 10] - [(1/8)x + 5] - [(1/16)x + 2.5] - [(1/32)x + 1.25] = 0
And....{with a little help from WolframAlpha}......x = 1240 coins {pounds ???}
Proof.....
The first thief gets 640 coins
The second thief gets 320 coins
The third theif gets 160 coins
The fourth thief gets 80 coins
And the last thief gets 40
So .... 640 + 320 + 160 + 80 + 40 = 1240 ......yep...that seems correct....
Thanks to Alan, I spotted an earlier mistake.....we both got the same answer approaching from "different poles"......
+870 Sorry, I was too busy to answer back to you.
OK, you both have
$$\textcolor[rgb]{1,0,0}{20/20}$$
And one cookie for each:
