There is a set containing $1, 2, \cdots, 199$, we proceed to continuously select (at random) $2$ numbers ($a$ and $b$) then compute the number $c = (a+b+ab)$. We remove both $a$ and $b$ while we add $c$ to this set. Finally what will the final remaining number be?
The question seems to suggest that no matter which two numbers you choose, in every possible ordering, the final answer will be the same number. Lets generalize the question for integers $1, 2, \cdots, n$.
For $n=1$, the answer is $1$. For $n = 2$, the answer is $(1+2+2) = 5$. For $n = 3$, the first iteration yields $(5,3)$, $(11,1)$, or $(2,7)$. Which all lead to $23$.
Initially the sum of the $n$ numbers is $\frac{n(n+1)}{2}$. In each iteration we remove $a$ and $b$, decreasing the sum by $a$ and $b$ while we add $(a+b+ab)$.
So the net change in the sum after each iteration will be $(-a+-b+a+b+ab )= ab$. So $Sum_{new} = Sum_{old} + ab$.
