A Spiral Thing. (CEMC)

Since the segment length increase by 1 every 2 segments, the total length after drawing even number $2n$ segments is given $\displaystyle L = \sum_{k=1}^n 2k = n(n+1)$ . We need to find the $n$ such that $n(n+1) < 3000$ . We can assume that $n = \lfloor \sqrt{3000} \rfloor = \lfloor 54.772 \rfloor = 54$ . We have $n(n+1) = 54\times 55 = 2970$ . This means that the length of the last segment is $3000-2970 = 30$ . This is shorter than the two segment before last which is 54. Therefore, the longest segment that Dana draws is $\boxed{54}$ .

The sum of numbers from 1 to $n$ is $\frac12n(n+1)$ . Sum of this series is double that, or $n(n+1)$ .

The pen died sometime during a stroke, not necessarily at the end of it, so we are looking into which $n$ would have the value $n(n+1)$ as close to 3000 as possible but still below it.

One way to do it is so solver $n^2+n=3000$ , which gives the answer $n=54.27$ , and round it down to a whole number $n=\boxed{54}$