Problem 3. A four-digit number $X$ is not divisible by 10. The sum of the number $X$ and the number written with the same digits in reverse order is equal to $N$. It turns out that the number $N$ is divisible by 100. Find $N$.
Answer: 11000.

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

According to the condition, $X+Y$ is divisible by 100, i.e., $1001(a+d)+110(b+c) \vdots 100$.

We have $1001(a+d) \vdots 10$, i.e., $a+d \vdots 10$, from which, since $a$ and $d$ are digits and $a \neq 0, 1 \leq a+d \leq 18$, thus $a+d=10$. Further, $1001 \cdot 10+110(b+c) \vdots 100$, i.e., $b+c+1 \vdots 10$, from which, since $b$ and $c$ are digits, $1 \leq b+c+1 \leq 19$, thus, $b+c=9$.

Therefore, $N=X+Y=1001 \cdot 10+110 \cdot 9=11000$.