2. Determine all quadruples of different two-digit natural numbers, for which the following conditions are simultaneously satisfied:

(i) The sum of the numbers in the quadruple that contain the digit 2 is 80.

(ii) The sum of the numbers in the quadruple that contain the digit 3 is 90.

(iii) The sum of the numbers in the quadruple that contain the digit 5 is 60.

(Jaroslav Zhouf)
SOLUTION. Consider any suitable quadruple. First, note that none of the three described sums can be obtained as the "sum" of a single number, since the number 80 does not contain the digit 2, the number 90 does not contain the digit 3, and the number 60 does not contain the digit 5.

According to the problem, the numbers containing the digit 5 have a sum of 60. We have already justified that this must be the sum of at least two numbers. However, it cannot be the sum of more than two numbers because

- none of them can have the digit 5 in the tens place, otherwise we would have a sum of at least $50+15$, which is more than 60;
- conversely, if all the numbers added had the digit 5 in the units place, their sum would have to be at least $15+25+35$, which is also more than 60.

Thus, the sum of 60 must be the sum of exactly two numbers, both of which have the digit 5 in the units place. These two numbers are therefore either 15 and 45, or 25 and 35. We will show that the first option is excluded.

Suppose, then, that our quadruple contains (the first two) numbers 15 and 45. Neither of them contains the digit 2, so the required sum of 80 must be the sum of the remaining two numbers. Similarly, due to the digit 3, we conclude that the sum of the remaining two numbers must be not 80, but 90. The obtained contradiction shows that the quadruple must necessarily contain the other two numbers 25 and 35.

Now notice that due to the numbers 25 and 35, our sums of 80 and 90 are not the sums of exactly two numbers, because the other addends would have to be equal to $80-25$, respectively $90-35$, i.e., 55 in both cases. Then, however, the sum of 60 of all numbers with the digit 5 (which we already know is achieved as $25+35$) would not hold.

Thus, the sum of 80 as well as the sum of 90 must necessarily be the sum of three numbers from our quadruple of four numbers; four numbers cannot be because of the numbers 25 and 35. The remaining two numbers therefore both contain the digit 2 and the digit 3. They must therefore be the numbers 23 and 32, so our quadruple must consist of the numbers $25, 35, 23, 32$. However, it is still necessary to verify that such a quadruple has the required sums of 80 and 90 (we already know that the sum $60=25+35$ holds). Indeed, $25+23+32=80$ and $35+23+32=90$.

Conclusion. The only suitable quadruple is formed by the numbers 25, 35, 23, 32.

NOTE. By modifying our solution process, we will show that the quadruple satisfying the problem statement remains unique even if we do not require that it be composed of distinct numbers.

The assumption of distinct numbers was first used in solving the question of whether the number 60 can be expressed as the sum of three or four numbers with the digit 5 in the units place. This can be easily ruled out even in the case where the numbers are not necessarily distinct. Indeed, the sum of three numbers ending in the digit 5 also ends in the digit 5, and the sum of four such numbers is equal to 60 only if they are the numbers 15, 15, 15, 15. However, such a quadruple does not satisfy the problem statement.

Second (and last) we used the assumption of distinct numbers in proving that the numbers 25, 35 must be supplemented by the numbers 23, 32. In the more general situation, it is necessary to rule out that the additional numbers are 23, 23 or 32, 32. This is obvious and the entire modified process is complete.

## GUIDING AND ADDITIONAL PROBLEMS:

N1. On the board are written three two-digit (not necessarily distinct) numbers such that the sum of those with the digit 1 is 36 and the sum of those with the digit 5 is 40. Determine these three numbers. [15, 21, 25. Focus on the given sum of 40. Since this number does not have the digit 5, it cannot be the "sum" of one or all three numbers written, which would have to end in the digit 5, as they are less than 50; it must therefore be the sum of two numbers ending in the digit 5, i.e., necessarily the numbers 15 and 25. The remaining third number must, together with the number 15, give the given sum of 36, so it is the number 21 (which indeed contains the digit 1).]

N2. On the board are written distinct two-digit numbers such that each of them contains the digit 5 and their sum is 75. Determine these numbers (find all possibilities). [Either one number 75, or two numbers 50 and 25, or three numbers 15, 25, and 35. If there is one number, then it is the number 75. If there are two numbers, then the digit 5 can only start one number and end one number, so we have $75=5 \star+\star 5$, which is only $50+25$. If there are at least three numbers, then the sum of all is at least $15+25+35=75$; they are therefore exactly three numbers, and those are 15, 25, and 35.]

D1. On the board are written 18 distinct two-digit numbers. The sum of those containing the digit 1 is 593. Determine all possible values of the sum of those numbers containing the digit 2. [The only value is 33. The number of all two-digit numbers containing the digit 1 is equal to the number 18 from the problem statement, they are the numbers $10,11, \ldots, 19,21,31, \ldots, 91$ and their sum is exactly the number 593 from the problem statement. Therefore, all these numbers are on the board and no others. Those containing the digit 2 are exactly 12 and 21.]

D2. Find all four-digit numbers $\overline{a b c d}$ with a digit sum of 12 such that $\overline{a b}-\overline{c d}=1$. $[69-\mathrm{C}-\mathrm{I}-1]$

D3. Find the smallest four-digit number $\overline{a b c d}$ such that the difference $(\overline{a b})^{2}-(\overline{c d})^{2}$ is a three-digit number written with three identical digits. [67-C-I-1]

D4. From the digits 0 to 9, 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. [70-B-I-1]