19. Find the last two digits of the sum

$$
1^{2}+2^{2}+\ldots+50^{2}-51^{2}-\ldots-100^{2}+101^{2}+\ldots 150^{2}-151^{2}-\ldots 200^{2}+\ldots-2000^{2}+2001^{2}+\ldots+2017^{2}
$$

(i.e., 50 numbers with a plus sign, 50 with a minus sign, and so on.)
# Answer: 85

Solution: Note that the numbers $\mathrm{n}^{2}$ and $(\mathrm{n}+50)^{2}$ give the same remainder when divided by 100. Therefore, in each hundred, the sum will end in two zeros. The last digits of the squares from 2001 to 2017 are:

| 01 | 04 | 09 | 16 | 25 | 36 | 49 | 64 | 81 | 00 | 21 | 44 | 69 | 96 | 25 | 56 | 89 , which in sum |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |

gives 685.