$a_{n}$ is the last digit of $1+2+\ldots+n$. Find $a_{1}+a_{2}+\ldots+a_{1992}$.
It is easy to compile the following table, from which we see that $\mathrm{a}_{\mathrm{n}}$ is periodic with period 20, and indeed the sum for each decade (from 0 to 9) is 35. Thus the sum for 1992 is $199.35+5+$ $6+8=6984$.

| $\mathrm{n}$ |  | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 17 | 18 | 19 | 20 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
| $a_{\mathrm{n}}$ |  | 0 | 1 | 3 | 6 | 0 | 5 | 1 | 8 | 6 | 5 | 5 | 6 | 8 | 1 | 5 | 0 | 6 |
| 3 | 1 | 0 | 0 |  |  |  |  |  |  |  |  |  | 6 |  | 1 | 5 |  |  |
| sum |  | 0 | 1 | 4 | 10 | 10 | 15 | 16 | 24 | 30 | 35 | 40 | 46 | 54 | 55 | 60 | 60 | 66 |
| 69 | 70 | 70 | 70 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |