Find the lowest sum

The sequence of consecutive integers is an arithmetic progression with common difference of 1. The sum of 50 consecutive integers is given by $S = \dfrac {50}2(2a + 49)$ , where $a$ is the first term. We note that $S$ increases with $a$ . For $S > 0$ , we have:

$\begin{aligned} \frac {50}2(2a+49) & > 0 \\ 50a + 1225 & > 0 \\ a & > - 24.5 \\ \implies a & \ge - 24 \end{aligned}$

Therefore for $S >0$ , the smallest $a=-24$ and the smallest $S = 25(2(-24)+49) = \boxed{25}$ .

Because our product must be zero, 0 must be one of the numbers in the range of (x) through (x+49).
Therefore, x in this case must be a number between -49 and 0.

We are then asked to find the sum of those numbers which can be written as such
An alternative way of writing this is 50x+1225, and we know the value of that must be greater than 0 (a positive number)
So we can set it up as 50x+1225>0
50x > -1225
x > -1225/50
x > -24.5
The first number x greater than -24.5 is -24. Therefore -24 is our starting number, x.

If we plug that back in to 50(-24)+1225 we get our answer, 25