## Task 5 - 330835

A six-digit natural number is called a "mirror number" if its first digit is equal to its sixth digit, its second digit is equal to its fifth digit, and its third digit is equal to its fourth digit.

a) Determine all "mirror numbers" that contain two digits 2, two digits 5, and two digits 7! Determine the sum $s$ of all these "mirror numbers"! What is the largest proper divisor of $s$?

b) Prove that for any three digits $a, b, c$ of which no two are the same, the following statement holds!

The sum of all "mirror numbers" that contain two digits $a$, two digits $b$, and two digits $c$ is divisible by 111111.

Hint: A proper divisor of $s$ is any divisor of $s$ that is (positive and) smaller than $s$.
a) For the arrangement of the three digits 2, 5, 7, there are exactly six possibilities; accordingly, all "mirror numbers" of the mentioned type are the numbers 257752, 275572, 527725, 572275, 725527, and 752257. Their sum is $s=3111108$.

The smallest natural number that is greater than 1 and a divisor of $s$ is 2. Therefore, the number $\frac{s}{2}=1555554$ is a divisor of $s$, smaller than $s$, and at the same time the largest number with these two properties, thus the largest proper divisor of $s$.

b) Correspondingly, for any three digits $a, b, c$, none of which are equal to each other:

To form the sum of the "mirror numbers" mentioned in the problem text using the usual addition method, one has to form the number $z=2 \cdot a+2 \cdot b+2 \cdot c$ at all six positions (units, tens, ..., hundred-thousands place) when adding. Therefore, this sum is

$$
z+10 \cdot z+10^{2} \cdot z+10^{3} \cdot z+10^{4} \cdot z+10^{5} \cdot z=111111 \cdot z
$$

and is thus divisible by 111111.