The numbers $1,2,...,100$ are written in a board. We are allowed to choose any two numbers from the board $a,b$ to delete them and replace on the board the number $a+b-1$.
What are the possible numbers u can get after $99$ consecutive operations of these?
Notice that the sum of the numbers on the board will decrease by 1 after 1 operation. After 99 consecutive operations of these, there is only 1 number left. And that number is $1+2+...+100-99=4951$
Note that the number of terms decreases by unity each time and that the sum of all terms decreases by one as well. Thus after $99$ consecutive applications, only one number could be left and that is $1+2+\cdots+100-99 = 4951$.
This, will be the same. Because, you will eventually add up all the integers. Thus, the only possible value, is $1+2+3+...+100-1\cdot 99=\boxed{4951}.$
lol what a joke
Note that the sum of the numbers decreases by $1$ each operation so the final number can only be the final sum, i. e. $1 + 2+ \dots + 100 -99 = 5050-99 = \boxed{4951}$.