1. We have two two-digit integers, which do not start with a 0. If you add these numbers together, you get the number $S$. If you swap the two digits of both numbers and add these two new numbers together, you get $4S$.

Determine all possible pairs of two-digit numbers for which this is true. Clearly indicate in your answer which two numbers belong together.
B1. $\{14,19\},\{15,18\}$ and $\{16,17\}$ We call the digits of our two numbers $a, b, c$ and $d$, so that the first number is equal to $10 a+b$ and the second number is equal to $10 c+d$. We thus get that

$$
S=10(a+c)+b+d
$$

The numbers we get by swapping the digits are $10 b+a$ and $10 d+c$. When we add these, we get

$$
4 S=10(b+d)+a+c
$$

If we now multiply the first equation by -4 and add it to the second equation, we get

$$
0=-39(a+c)+6(b+d), \quad \text { or } \quad 13(a+c)=2(b+d)
$$

The right side of the last equation can only be divisible by 13 if $b+d$ is divisible by 13. But $b+d$ is at least 0 and at most 18, so this can only be if $b+d=0$ or $b+d=13$.

If $b+d=0$, then $a+c=0$ must also hold. Both numbers would then start with a 0, which was not allowed. This possibility is therefore ruled out.

If $b+d=13$, then $a+c=2$ must hold. The digits $a$ and $c$ cannot be zero. So we have $a=c=1$. For the pair $\{b, d\}$, we have the possibilities $\{4,9\},\{5,8\}$ and $\{6,7\}$. The possible starting pairs are thus $\{14,19\},\{15,18\}$ and $\{16,17\}$. It is easy to check that these are indeed solutions. In all cases, we get the sum $S=33$ and the sum of the reversed numbers is exactly $132=4 S$.