6. For the numbers $1000^{2}, 1001^{2}, 1002^{2}, \ldots$, the last three digits are discarded. How many of the first terms of the resulting sequence form an arithmetic progression?
Answer: 32.

Solution. $(1000+k)^{2}=1000000+2000 k+k^{2}$. As long as $k^{2}<1000$, after discarding the last three digits, the number $1000+2 k$ will remain, i.e., each term of the sequence will be 2 more than the previous one. When $k=31: k^{2}=961<1000$, and when $k=32: k^{2}=1024$. Therefore, the element obtained from $1032^{2}$ will differ from the previous one by 3. Thus, the length of the progression is the number of numbers from 1000 to 1031, i.e., 32.