1. Using the digits 0 to 9, we form two-digit numbers $A B, C D, E F, G H, I J$, using each digit exactly once. Determine how many different values the sum $A B+C D+E F+G H+I J$ can take and what these values are. (Expressions like 07 are not considered two-digit numbers.)

(Jaroslav Zhouf)
SOLUTION. The value of the sum being examined

$$
S=A B+C D+E F+G H+I J
$$

depends only on which five digits are in the tens place of the formed numbers, and which five are in the units place. If we determine the sums of the numbers in both of these sets of five

$$
S_{1}=A+C+E+G+I \quad \text { and } \quad S_{0}=B+D+F+H+J
$$

we will clearly have $S=10 S_{1}+S_{0}$. According to the problem statement, the addends in both sums $S_{1}$ and $S_{0}$ together form all 10 digits from 0 to 9. Therefore, the sum $S_{1}+S_{0}$ is the same as the sum $0+1+\ldots+9$, which equals 45. Thus, we have $S_{0}=45-S_{1}$, and for the sum $S$ we get the expression

$$
S=10 S_{1}+S_{0}=10 S_{1}+\left(45-S_{1}\right)=9 S_{1}+45=9 \cdot\left(S_{1}+5\right) .
$$

For the solution of our problem, it is sufficient according to the last formula to determine what values the sum $S_{1}$ can take. This is the sum of the digits $A, C, E, G, I$ in the tens place of five two-digit numbers, and so according to the problem statement, it can be the sum of any five different non-zero digits - the remaining five unrepresented digits (including zero) can then be used to arbitrarily fill the units places of the formed numbers. For the sum of such five digits, it is clearly true that

$$
15=1+2+3+4+5 \leqq A+C+E+G+I \leqq 5+6+7+8+9=35 .
$$

We will convince ourselves that the sum $S_{1}=A+C+E+G+I$ can take all 21 values in the range from 15 to 35. For example, the value 16 is obtained by changing the largest number 5 in the set $1,2,3,4,5$ to the number 6. We can continue increasing the sum $S_{1}$ by 1 in this way until we get the set $1,2,3,4,9$. Then we start increasing the number 4 by one repeatedly until we get the set $1,2,3,8,9$. We proceed similarly further until we finally reach the set $5,6,7,8,9$. In this way, we indeed get every integer value of the sum $S_{1}$ in the range from 15 to 35 (of which there are 21).

Thanks to the last finding, according to the previously derived formula $S=9 \cdot\left(S_{1}+5\right)$, we conclude that the sum $A B+C D+E F+G H+I J$ can take exactly 21 values, which are multiples of nine in the range from 180 to 360 (the boundary values we calculated as $9 \cdot(15+5)=180$ and $9 \cdot(35+5)=360)$.

SUPPLEMENTARY AND ADDITIONAL PROBLEMS:

N1. From two different digits $A, B$ we form two-digit numbers $A B$ and $B A$. Prove that their difference is divisible by 9. $[A B-B A=(10 \cdot A+B)-(10 \cdot B+A)=9(A-B)$.]

N2. We "reduce" the situation from the competition problem. From the digits $1,2,3,4$ we form two-digit numbers $A B$, $C D$, using each digit exactly once.

a) Find the smallest and the largest possible value of $A B+C D$.

b) Determine the smallest possible positive difference between two such formed numbers $A B+C D$.

a) The minimum is 37, the maximum is 73. The minimum, respectively the maximum, is obtained when we place the smallest digits 1 and 2 in the tens place, respectively in the units place. b) 9. Each formed number $A B+C D$ gives the same remainder as the number $A+B+C+D$ equal to $1+2+3+4=10$, which gives a remainder of 1. Therefore, the difference between any two formed numbers $A B+C D$ is divisible by nine, so the sought minimum is a positive multiple of the number 9. That it is no more than 9 is shown by the example $(12+43)-(12+34)=9$.]

N3. From the numbers 1, 2, 3, 4, 5, 6 we select three different ones. Prove that their sum can take any integer value from 6 to 15. [Finding the smallest and largest value is not enough. It is necessary to describe how to construct any permissible sum, for example 13. The description can be an algorithm that goes through all sums from 6 to 15. Starting with the triplet $1,2,3$, we can repeatedly increase the number 3 by one until we get the triplet 1, 2, 6. Then we start increasing the number 2 until we get $1,5,6$. Finally, we increase the number 1 until we get 4, 5, 6, thus covering all numbers from 6 to 15.]

D1. Honza has three cards, each with a different non-zero digit. The sum of all three-digit numbers that can be formed from these cards is a number 6 greater than three times one of them. What digits are on the cards? [61-C-II-2]

D2. Find all eight-digit numbers such that by erasing some quartet of adjacent digits, we get a four-digit number that is 2019 times smaller. [68-B-I-1]