Just to put yorksnbeans out of her misery...

Have a look back at the question if you missed it.

For starters let's call the number of gold coins they find p for purse, and let the wealth of each of the three merchants be x, y and z gold coins.

So from the problem we're told

x + p = 2(y + z)and we have to find values which satisfy all three of those equations.

y + p = 3(x + z)

z + p = 5(x + y)

First write those sums in terms of the purse

p = 2y + 2z - x --- (1)----------------------------------------

p = 3x + 3z - y --- (2)

p = 5x + 5y - z --- (3)

Let's get rid of the purse for now by taking (3) minus (2)

p - p = 5x - 3x + 5y + y - z - 3zadd up the like terms

2x + 6y - 4z = 0and divide by 2 for simplicity

x + 3y - 2z = 0 --- (A)-----------------------------------------

Do the same again with (3) minus (1)

p - p = 5x + x + 5y - 2y - z - 2zso:

6x + 3y - 3z = 0This time divide by 3 to make it simpler

2x + y - z = 0 --- (B)-----------------------------------------

Now, 2 times (B) minus (A) gives

4x + 2y - 2z - x - 3y + 2z = 0so

3x - y = 0or

y = 3x --- (C)and 3 times (B) minus (A) gives

6x + 3y - 3z -x -3y + 2z = 0so

5x - z = 0or

z = 5x --- (D)------------------------------------------

Gold coins come in whole numbers (usually), the smallest whole number is 1.

Looking at (C) and (D) tells us that x has the least gold, so let x = 1. Then (C) and (D) give values for y and z, and plugging those values into (1), (2) or (3) gives us the purse. Result!

x = 1Of course x could be any whole number, there is no unique solution, ie:

y = 3

z = 5

p = 15

Great fun! Thanks to Mr P at the bar for setting it. I'll look out for more little posers like this, it keeps the grey cells active.

## 2 comments:

Holy Crap! No wonder I didn't get it! :-)

It looks much worse than it is yorksnbeans, honest.

I'm about to post another which may be a bit simpler, no x's or y's needed. Just numbers and a bit of lateral thinking.

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