3. Each three-digit number has the expression $m=100 a+10 b+c$, where $a, b, c, a \neq 0$, are its digits. The three-digit number with the same digits in reverse order then has the expression $m^{\prime}=100 c+10 b+a, c \neq 0$. Since the order of the numbers $m$ and $m^{\prime}$ is not considered, for definiteness, let's assume that $m \leqq m^{\prime}$ or $a \leqq c$, where $a, c \in\{1,2,3, \ldots, 9\}$ and $b \in\{0,1,2, \ldots, 9\}$.

For the magic number $x$, according to the notation of the digits, we have

$$
x=m+m^{\prime}=101(a+c)+20 b .
$$

We see that the value of $x$ depends less on the individual digits $a$ and $c$ than on their sum $s=a+c$, which can take values $s \in\{2,3, \ldots, 18\}$. We will work further only with the expression $x=101 s+20 b$.

Suppose for a moment that some magic number $x$ can be expressed as the sum $101 s+20 b$ in two different ways:

$$
x=101 s+20 b=101 s^{\prime}+20 b^{\prime} .
$$

From the equality $101\left(s-s^{\prime}\right)=20\left(b-b^{\prime}\right)$ and the coprimality of the numbers 101 and 20, it follows that the number 101 must divide the number $b-b^{\prime}$. However, since $b$ and $b^{\prime}$ are digits, we have $-9 \leqq b-b^{\prime} \leqq 9$. In this interval, we find only one number divisible by 101, and that is the number 0. Therefore, $b-b^{\prime}=0$ or $b=b^{\prime}$, and thus $s=s^{\prime}$. This contradicts the assumption that the number $x$ has two different expressions of the form (1). This means that in the expression $x=101 s+20 b$, each magic number $x$ has a uniquely determined digit $b$ and a uniquely determined sum $s$.

The number of ways in which a magic number can be expressed as the sum $m+m^{\prime}$ or $101 s+20 b$ is therefore equal to the number of ways in which the corresponding value of $s$ can be expressed as the sum of two digits $a$ and $c$, where $1 \leqq a \leqq c \leqq 9$. In the set $\{2,3, \ldots, 18\}$, the number $s=10$ has the most such expressions, which can be expressed by exactly five valid sums:

$$
10=1+9=2+8=3+7=4+6=5+5 \text {. }
$$

Indeed: in the case $s \leqq 9$, from the inequality $a+c \leqq 9$ and the assumption $a \leqq c$, it follows that $a \leqq 4$, so the smaller digit $a$ can take at most four values, just as the larger digit $c$ in the case $s \geqq 11$, where from the relations $a+c \geqq 11$ and $a \leqq c$ it follows that $c \geqq 6$.

The most (five) sums $m+m^{\prime}$ can be expressed by magic numbers of the form $101 \cdot 10+20 b$, where $b \in\{0,1,2, \ldots, 9\}$, which are the numbers in the ten-element set

$$
\{1010,1030,1050, \ldots, 1190\}
$$

For a complete solution, award 6 points, of which 3 points for the proof that a magic number has a unique expression as the sum $101 s+20 b$. The observation that the number $s=10$ has the most expressions $s=a+c$ is so obvious that it can be stated without justification. In the case of a correct procedure, deduct no more than 1 point for numerically incorrect calculation of some of the 10 solutions.
3. We will call a number magical if it can be expressed as the sum of a three-digit number $m$ and a three-digit number $m^{\prime}$ written with the same digits in reverse order. Some magical numbers can be expressed in this way in multiple ways; for example, $1554=579+975=777+777$. Determine all magical numbers that have the most such expressions $m+m^{\prime}$. (The order of $m$ and $m^{\prime}$ is not considered.)

The written part of the school round in category B takes place

## on Thursday, January 22, 2009

so that it starts in the morning and the contestants have 4 hours of pure time to solve the problems. For each problem, a contestant can earn 6 points, and a successful solver is a student who scores 10 points or more. Allowed aids are writing and drawing supplies, school MF tables, and calculators without a graphical display. These details will be communicated to the students before the start of the competition.