$(a_{1}, a_{2},...,a_{20}) = (1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0)$, so $a_{i}= a_{20+i}$. $\Sigma_{i=1}^{20}a_{i}= 70$, so $\Sigma_{i=1}^{1992}a_{i}= \Sigma_{i=1}^{2000}a_{i}-\Sigma_{i=13}^{20}a_{i}= 7000-16=6984$.

Jutaro wrote: For every positive integer $n$ we define $a_{n}$ as the last digit of the sum $1+2+\cdots+n$. Compute $a_{1}+a_{2}+\cdots+a_{1992}$. Let $S_{n}=1+2+\cdots+n=\frac{n(n+1)}{2}$. It's immediate to see that $S_{n+20}=S_{n}$ (mod $10$). So $a_{1}+a_{2}+\cdots+a_{1992}=99(a_{1}+a_{2}+\cdots+a_{20})+a_{1}+a_{2}+\cdots+a_{12}=6984$. -- Patrick

We try to find a pattern by listing the last digit of the first sums, and notice that $a_{20}=0$, so beyond that point the cycle will repeat every 20 numbers and the sum of each cycle is equal to 70, thus $a_1 +a_2+...a_{1980}=99 \cdot 70$ adding the last 19 numbers we get that the sum is equal to $99 \cdot 70 +54=6984$

Also 2017 AIME I #3 except slightly different.

I participated at this event in Venezuela competing on behalf of Puerto Rico. A lot of people got a score of 8 out of 10 on this problem including myself because although my answer was correct, they took 2 points of because like all of the solutions posted here, merely showing that the pattern repeats after 20 was not enough. "see" that the pattern repeats, or explaining that it repeats after 20 is not sufficient proof according to the graders. All of the answers provided above would have also gotten 8 out of 10. I've included the proof here: https://artofproblemsolving.com/wiki/index.php/1992_OIM_Problems/Problem_1