3. Determine the smallest natural number that has the property of being reduced 57 times if its first digit from the left is erased.
Solution. Let's denote the first digit of the desired number by $a$, and the number formed by the remaining digits by $b$. Then $10^{k} a + b = 57 b$ where $k$ is a natural number, i.e., $10^{k} a = 56 b$, or $10^{k} a = 7 \cdot 8 b$. However, $\mathrm{GCD}(10,7)=1$, so $7 \mid a$ and since $a$ is a digit, we get $a=7$. Now $10^{k} = 8 b$. The number $10^{k}$ must be divisible by 8, so from this $k \geq 3$, and since we are looking for the smallest number, we get $k=3$ and $b=125$. Finally, the desired number is 7125.