[Guest]

A rich merchant had collected

many gold coins. He did not

want anybody to know about

them. One day, his wife asked,

"How many gold coins do we

have?"

After pausing a moment, he

replied, "Well! If I divide the coins

into two unequal numbers, then

50 times the difference

between the two numbers

equals the difference between

the squares of the two

numbers."

The wife looked puzzled. Can

you help the merchant's wife by

finding out how many gold coins

they have?

[bonanova]

50

[a+b][a-b]=a²-b²

[Guest]

Normally avoiding math

50 (A - B) = (A x A) - (B x B).

I usually can't explain as I need to do sums in my head, or in code

[Guest]

Let x, y be two unique groups of coins. x - y != 0, and assume x > y.

50 * (x - y) = x^2 - y^2 <-----------set up problem

50 * (x - y) = (x + y)(x - y) <------factor the binomial

50 = (x + y) <-----------------------cancel (x - y) from both sides

50 = answer

[Guest]

50.

50x-50y=x^2-y^2

50(x-y)=(x+y)(x-y) factor

50=x+y divide by common factor