The solution to yesterday's math problem

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

and we have to find values which satisfy all three of those equations.

First write those sums in terms of the purse

p = 2y + 2z - x         --- (1)
p = 3x + 3z - y         --- (2)
p = 5x + 5y - z         --- (3)

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

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

add up the like terms

2x + 6y - 4z = 0

and divide by 2 for simplicity

x + 3y - 2z = 0         --- (A)

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

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

so:

6x + 3y - 3z = 0

This time divide by 3 to make it simpler

2x + y - z = 0         --- (B)

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

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

so

3x - y = 0

or

y = 3x         --- (C)

and 3 times (B) minus (A) gives

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

so

5x - z = 0

or

z = 5x         --- (D)

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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 = 1
y = 3
z = 5
p = 15

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

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.