A six-digit number is "prized" if the sum of its first three digits is equal to the sum of its last three digits. For example, 342531 is prized because $3+4+2=5+3+1$.

(a) What is the largest and the smallest prized number, with 6 different digits?

(b) Show that the sum of all prized numbers, with 6 different digits, is divisible by 13.
(a) The largest winning number must start with 98. Thus, the number we are looking for is of the form: $98 a b c d$. By hypothesis, we have: $9+8+a=b+c+d$. To maximize $a$, we need $b+c+d$ to be maximized, and this occurs when $b=7, c=6$, and $d=5$. In this case, $a=1$, and consequently, the largest winning number is 981765.

Let's now determine the smallest winning number. Let's try a number of the form $10 a b c d$. It is not difficult to verify that 108234 is the smallest winning number.

(b) If the number $A B C D E F$ is a winning number, then the number $D E F A B C$ is also a winning number. The sum of these numbers is:

$$
\begin{aligned}
A B C D E F+D E F A B C & =(1000 A B C+D E F)+(1000 D E F+A B C) \\
& =1001(A B C+D E F)=13 \times 11 \times 7 \times(A B C+D E F)
\end{aligned}
$$

By summing all the winning numbers with 6 different digits in pairs, it results that each pair is divisible by 13. Therefore, the sum of all of them is divisible by 13.

Note: In fact, it is also divisible by 11 and 7.