## Task 17/67

A four-digit number (in the decimal system) is sought, for which the following is known:

1. The sum of all two-digit numbers that can be formed from any two digits of the sought number is 594.
2. If the sought number is divided by its cross sum, the result is a number that is equal to seven times the last digit of the sought number.
3. All digits of the sought number are different from each other and not equal to zero.
Let $z=1000 a+100 b+10 c+d$ be the desired number, $Q=a+b+c+d$ its digit sum. The two-digit numbers that can be formed from the digits $a, b, c, d$ are

$10 a+b, 10 a+c, 10 a+d, 10 b+a, 10 b+c, 10 b+c, 10 c+a, 10 c+b, 10 c+d, 10 d+a, 10 d+b, 10 d+c$

The sum of all these two-digit numbers is $33(a+b+c+d)=594$, so $a+b+c+d=18=Q$. From $\frac{z}{Q}=17 d$ it follows immediately that $z=Q \cdot 17 d=306 d$. Thus, we get the equation

$$
z=1000 a+100 b+10 c+d=306 d \quad ; \quad 10(100 a+10 b+c)=305 d
$$

The number $d$ must therefore be even:

$d=2$, then $z=612$, which is excluded because it is not four digits

$d=4$, then $z=1224$, which is excluded because $b=c$

$d=6$, then $z=1836$, possible, verification confirms correctness

$d=8$, then $z=2448$, which is excluded because $b=c$

The number 1836 is thus the only solution to the problem.