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# The unfortunate merchant

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A merchant returns home, perky. His coin purse is full to bursting.
Suddenly, a man flows on him, and tells him:

"If you want to stay alive,
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

EinsteinJr  May 17, 2015

#2
+26399
+8

I've tackled it from the other end:

.

Alan  May 17, 2015
Sort:

#1
+80983
+8

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"......

CPhill  May 17, 2015
#2
+26399
+8

I've tackled it from the other end:

.

Alan  May 17, 2015
#3
+886
+5

Sorry, I was too busy to answer back to you.

OK, you both have

\$\$\textcolor[rgb]{1,0,0}{20/20}\$\$

EinsteinJr  May 19, 2015

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