B1. Every three-digit number can be written in reverse. 176 can be reversed to 671.

Determine all three-digit numbers that are equal to 3 times their reversed plus the sum of their digits.

In formula: $\overline{d c}=3 \times \overline{d a}+(a+b+c)$
B1. $100 a+10 b+c=300 c+30 b+3 a+a+b+c$ thus $300 c+21 b=96 a$ or $100 c+7 b=32 a$ with $a$, $b$, and $c$ being integers and $a, b, c \leq 9$.

Since 100 and 32 are both multiples of four, $b$ must also be a multiple of four. Therefore, the problem reduces to finding solutions for $a$ and $c$ for $b=0$, $b=4$, and $b=8$.

- $b=0$, then $100 c=32 a$ so $25 c=8 a$ and thus $a=25,50, \ldots$ which gives no solutions.
- $b=4$, then $25 c+7=8 a$ or $8 a-7=25 c$. Only $a=4, c=1$ provides a solution.
- $b=8$, then $8 a-14=25 c$. Only $a=8, c=2$ provides a solution.

In total, there are thus two solutions, 441 and 882.