Galochkina A.i.

The digits 1, 2, 3,..., 9 are arranged in a circle in some arbitrary order. Every three consecutive digits, when read clockwise, form a three-digit number. Find the sum of all nine such numbers. Does this sum depend on the order in which the digits are arranged?

#
In our numbers, each digit appears exactly once in each of the hundreds, tens, and units places.

## Solution

A three-digit number, where the hundreds place is digit $a$, the tens place is digit $b$, and the units place is digit $c$, is equal to $100a + 10b + c$. (For example, $394 = 3 \cdot 100 + 9 \cdot 10 + 4$.) Looking through our nine three-digit numbers in a cycle, we notice that each digit appears exactly once in each of the hundreds, tens, and units places. This means that each digit will enter our sum once with a coefficient of 100, once with a coefficient of 10, and once with a coefficient of 1. Therefore, the desired sum does not depend on the order in which the digits are written and is equal to

$$
(100 + 10 + 1)(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 111 \cdot 45 = 4995
$$

## Answer

4995; does not depend.