10.1. Find all three-digit natural numbers $A$, the square of which ends in $A$.
Answer: $A=376,625$.

Solution. By the condition, $A^{2}-A=A(A-1)$ is divisible by $1000=2^{3} \cdot 5^{3}$. Due to the mutual simplicity of $A$ and $A-1$, one of them is divisible by $2^{3}=8$, and the other by $5^{3}=125$.

If $A$ is divisible by 125, then $A \in\{125,250,375,500,625,750,875\}$ and $A-1 \in\{124,249,374,499,624,749,874\}$ - is divisible by 8, which is only possible when $A=625$.

If $A-1$ is divisible by 125, then $A \in\{126,251,376,501,626,751,876\}$ is divisible by 8, which is only possible for $A=376$. Indeed, $376^{2}=141376$.

Criteria. In the statement that one of $A$ and $A-1$ is divisible by $2^{3}=8$, and the other by $5^{3}=125$, there is no reference to their mutual simplicity: deduct 3 points.