1. Variant 1.

When multiplying two two-digit numbers, a four-digit number $A$ is obtained, where the first digit matches the second, and the second-to-last digit matches the last. Find the smallest $A$, given that $A$ is divisible by 51.
Answer: 1122.

Solution.

Notice that $A=\overline{x x y y}=x \cdot 11 \cdot 100+y \cdot 11=11 \cdot(100 x+y)$. Since 51 and 11 are coprime, then $100 x+y$ is divisible by 51. The minimum $x=1$, so $y=2$ (the only number from 100 to 109 divisible by 51 is 102).