60(1188). Find the smallest four-digit number which, after being multiplied by 21, becomes a square of a natural number.
Solution. Since $21=3 \cdot 7$ (3 and 7 are prime numbers), for the product $21 m$ (where $m$ is the desired number) to satisfy the condition of the problem, it is necessary that the equality $21 m=21 \cdot 21 a$ holds, where $a$ is the square of the smallest natural number such that $21 a$ is a four-digit number. By trial, we find $a: a=7^{2} ; m$ is found from the equality $m=21 a$.

Answer: 1029.

Remark. After solving this problem, it is advisable to propose a more general one: “The product $21 m$, where the factor $m$ is a four-digit number, can be represented as the square of a natural number. Find $m$.”

Answer: $m \in \left\{21 \cdot 7^{2} ; 21 \cdot 8^{2} ; 21 \cdot 9^{2} ; \ldots ; 21 \cdot 21^{2}\right\}$.