Problem 3. A four-digit number $X$ is not divisible by 10. The sum of the number $X$ and the number obtained from $X$ by swapping its second and third digits is divisible by 900. Find the remainder when the number $X$ is divided by 90.
Answer: 45.

Solution. Let $X=\overline{a b c d}=1000 a+100 b+10 c+d, Y=\overline{a c b d}=1000 a+100 c+10 b+d$, where $a$, $b, c, d$ are digits and $a \neq 0, d \neq 0$.

According to the condition, $X+Y$ is divisible by 900, i.e., $2000 a+110(b+c)+2 d \vdots 900$.

We have, $2 d \vdots 10$, i.e., $d \vdots 5$, so since $d \neq 0$ and $d$ is a digit, $d=5$. Next, $110(b+c)+10 \vdots 100$, i.e., $b+c+1 \vdots 10$, from which, since $b$ and $c$ are digits, $1 \leq b+c+1 \leq 19, b+c=9$. Finally, $2000 a+110 \cdot 9+10 \vdots 9$, i.e., $2 a+1 \vdots 9$, from which, since $a$ is a digit, $a=4$.

Thus, $X=4000+90 b+90+5=90 q+45$.