An algebra problem by A Former Brilliant Member

Suppose the coins are divided into two unequal parts $x$ and $y.$

According to the problem ,

$32(x-y) = (x^2-y^2)$

$32(x -y) = (x + y)(x-y)$

We know that $x \neq y.$

Therefore $(x-y) \neq 0.$ Hence we can cancel out $(x-y)$ term.

After cancelling out , we obtain

$(x+y)=32.$

Therefore the required answer is $\boxed{32}.$

Assume that $x$ and $y$ are the two unequal numbers as written in the question. Therefore, converting the words into an equation, this becomes:

$32 (x - y) = x^{2} - y^{2}$

Now to solve:

Move like terms together: $32 = \frac{x^{2} - y^{2}}{x - y}$

Factor the numerator: $32 = \frac{(x - y) (x + y)}{x - y}$

Reduce: $32 = x + y$

Therefore the answer is $32$ .

32 divided by 2 is equal 16,then 16 multiply by 4 which is 64,lastly divided 64 by 2,which is 32