$2 \cdot 51$ Remove some numbers from the set $1,2,3,4, \cdots, 1982$ so that in the remaining numbers, no number is equal to the product of any two other numbers. How many numbers must be removed at a minimum to achieve this? How can this be done?
［Solution］If 43 numbers: $2, 3, \cdots, 44$ are crossed out, then in the remaining numbers, no number is equal to the product of any other two numbers.

If fewer than 43 numbers are crossed out, then there is at least one triplet $(k, 89-k, k(89-k))$ consisting of numbers that have not been crossed out, where $2 \leqslant k \leqslant 43, 46 \leqslant 89-k \leqslant 87, k(89-k) \leqslant\left(\frac{89}{2}\right)^{2}=1980.25$.
2 - 52 The square of a certain positive integer has the last three digits as non-zero identical digits. Find the smallest positive integer with this property.
(Japan Mathematical Olympiad Team Selection Test, 1990)
[Solution] Let $p=10 a \pm b(a, b \in Z, 1 \leqslant b \leqslant 5)$, then
$$
p^{2} \equiv 10 \cdot (\pm 2 a b) + b^{2} (\bmod 100).
$$

It can be verified that when $b=1,3,4,5$, the tens and units digits of $p^{2}$ have opposite parity; only when $b=2$, the last two digits of $p^{2}$ have the same parity. Therefore, the required $p$ must be of the form $10 a \pm 2$, and $p=12, p^{2}=144$, the last two digits are 4.

By calculating $12^{2}, 18^{2}, 22^{2}, 28^{2}, 32^{2}, 38^{2}$, it is known that the smallest positive integer whose last three digits are non-zero and identical is 38.