B1. We rewrite the equation as $\mathrm{v} \overline{c d a b}+\overline{13 c d}=\overline{a b 20}$. Since $b$ and $d$ are non-zero digits, from the equation for the units place, we get $b+d=10$. We subtract this from both sides of the equation to get $\overline{c d a 0}+\overline{13 c 0}=\overline{a b 10}$, and then divide the equation by 10 to get $\overline{c d a}+\overline{13 c}=\overline{a b 1}$. Since $a$ and $c$ are non-zero digits, their sum cannot be 1, so from the equation for the units place, we get $a+c=11$. We subtract this from both sides of the equation and divide the equation by 10 to get $\overline{c d}+\overline{13}=\overline{a(b-1)}$, since $b-1 \geq 0$. We consider two cases. If $d \leq 6$, then $d+3=b-1$ and $c+1=a$. From all the equations, we calculate that $a=6, b=7, c=5$ and $d=3$. If $d \geq 7$, then $d+3=10+(b-1)$ and $c+1=a-1$. In this case, from the equations, we calculate $c=\frac{9}{2}$, which is a contradiction. The only solution to the equation is therefore $(a, b, c, d)=(6,7,5,3)$.

Identifying that $b+d=10$ 1 point

Identifying that $a+c=11$ 1 point

Writing the equation $\overline{c d}+\overline{13}=\overline{a(b-1)}$ 1 point

Considering the case $d \leq 6$ and writing the relationships between the digits ...................... 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_6340f9febbabfa1859cdg-25.jpg?height=51&width=1576&top_left_y=2373&top_left_x=274)
Considering the case $d \geq 7$ and identifying that there is no solution ..................... 1 point (If the contestant just writes the solution, they get 1 point.)
B1. Find all quadruples of non-zero digits $a, b$, $c$ and $d$, for which $\overline{a b 20}-\overline{13 c d}=\overline{c d a b}$.