1B. If the digit 1 is appended to the right of a three-digit number, a four-digit number is obtained which is 3 times larger than the number obtained by appending the digit 2 to the left of the given three-digit number. What is that three-digit number?
Solution. First method. Let the desired three-digit number be $\overline{a b c}$. From the condition of the problem, it follows that $\overline{a b c 1}=3 \cdot \overline{2 a b c}$. From the last equality, we get

$$
10 \overline{a b c}+1=3(2000+\overline{a b c}), \text { i.e. } 7 \overline{a b c}=5559
$$

Finally, $\overline{a b c}=857$.

Second method. Let the desired three-digit number be $\overline{a b c}$. From the condition of the problem, it follows that $\overline{a b c 1}=3 \cdot \overline{2 a b c}$. The only single-digit number that, when multiplied by 3, gives a units digit of 1 is the number 7, so $c=7$. Now, $\overline{a b 71}=3 \cdot \overline{2 a b 7}$. In the number on the left side, the tens digit is 7, so with a carry of 2, we get that the product of $b$ and 3 must give a units digit of 5. Therefore, $b=5$. Thus, $\overline{a 571}=3 \cdot \overline{2 a 57}$. Further, in the number on the left side, the hundreds digit is 5, so with a carry of 1, we get that the product of $a$ and 3 must give a units digit of 4. Therefore, $a=8$. Finally, $8571=3 \cdot 2857$.