A math riddle. Free points for grabs!?

Question

This is easy, yet most ppl dont get it......

A rich merchant had collected many rare gold coins. He did not want anybody to know about them. One day, one of his servants asked, "How many gold coins do you have, sir?"

He replied, "If I divide the coins into two unequal numbers, then 29 times the difference between the two numbers equals the difference between the squares of the two numbers."

How many does he have (this is too easy)

Answer 1

29(X-Y)=X2-Y2 then X+Y=?

X2-Y2=(X=Y)(X-Y)

29(X-Y)=(X+Y)(X-Y)

X+Y= 29

the rich merchant had 29 gold coins

Answer 2

29 is my best guess. The math way is this: If Z gold coins are divide into X and Y so that X+Y=Z, then:
29(X-Y)=X^2-Y^2 and everybody knows that the right side can be simplified so:
29(X-Y)=(X+Y)(X-Y). We can divide by (X-Y) cos' seeing as their unequal, their subtraction cannot be zero. So:
29=X+Y.
Yet I fail to comprehend the easy way out of this. Is there one?

Answer 3

If the two numbers are, x and y

29( x-y ) = x^2-y^2 ( assuming that x is the larger number)

29(x - y) = (x-y)(x+y)

29 = x + y

That is,
he has 29 gold coins.

Answer 4

9 and 10= 19 gold coins

Answer 5

He could have 29 gold coins
Divide them into piles of 0 and 29
(29-0)^2 = 29^2-0^2

Answer 6

My guess is it's a nonsensical reply. The squares of the the two numbers would far exceed 29 times each individual number. I am interested to see the other answers to this question.

Oh, never mind. You said the DIFFERENCE beween the squares of the two numbers. Still am curious about the right answer to this one.

P.S. I think "Math_kp did it. Smart guy -- I don't think this was "too easy."

Answer 7

Simplest sum,
if divided into 'a' and 'b'
29 * (a ~ b) = (a^2 ~ b^2)
so 29(a-b) = (a+b)(a-b)

so a+b = 29, i we divide anyhow like (0,29), (1,28)...(15,14) this condition will be satisfied

Proof: Simple proof, if we take any two numbers and square it, it will be product of sum of the number and difference of the numbers.
so as per the problem some x times difference of two numbers is difference of the square so naturally sum of the two numbers is x :)

Answer 8

Total coins is 29pcs.

You can solve by:
29(x-y)=x^2-y^2

Get the factor of x^2-y^2
29(x-y)=(x-y)(x+y)

cancel out (x-y)
29=x+y

Therefore the total coins are 29

Answer 9

Algebraically, as many of the others have solved:
29(x-y)=x^2-y^2
29(x-y)=(x-y)(x+y)
so (x+y)=29

Without just algebra, remember that Difference between two squares is same as product of their difference and sum.
eg. for numbers 9,999,975 and 25, mentally calculate the difference between their squares. Easiest way is (9,999,975-25)(9,999,975+25), right?
Same principle in this riddle. :)

Answer 10

suppose they are divided into x and y

29(x-y) = x^2-y^2

or 29(x-y) = (x+y)(x-y)
or x+y = 29(can divide by x-y as they are not same
so number of coins= x+y = 29

Answer 11

Well, Larry, when you have more variables than equations, it means there is more than one solution. And indeed, X and Y can be any numbers that add to 29.