## Task 2 - 250712

Determine all two-digit natural numbers that satisfy the following conditions:

(1) The difference between the two digits is 5.

(2) If the tens digit and the units digit are swapped, the resulting two-digit number is 4 more than double the original number.
The condition (1) is precisely met by the two-digit numbers listed in the first column of the following table, with the number 50 being immediately discarded since no two-digit number results from swapping its digits.

In the second column, the number resulting from swapping the digits is listed, in the third column, twice the number in the second column. In the fourth column, the original number (i.e., the one in the first column) increased by 4 is listed.

| 16 | 61 | 122 | 20 | 61 | 16 | 32 | 65 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 27 | 72 | 144 | 31 | 72 | 27 | 54 | 76 |
| 38 | 83 | 166 | 42 | 83 | 38 | 76 | 87 |
| 49 | 94 | 188 | 53 | 94 | 49 | 98 | 98 |

For the number 94 in the first column, equality is achieved in the third and fourth columns. Therefore, only the number 94 satisfies both conditions (1) and (2).

Alternative solution:

If \(a\) and \(b\) are the tens and units digits, respectively, of a number that meets the conditions of the problem, then:

The number is \(10a + b\), the number resulting from swapping the digits is \(10b + a\), and according to (2):

\[
2 \cdot (10b + a) = 10a + b + 4 \quad \Rightarrow \quad 19b = 4 \cdot (2a + 1)
\]

Since \(19b\) is the product of 4 and the odd number \(2a + 1\), and since 19 and 4 are coprime, it follows that \(b\) is divisible by 4 but not by 8. The only number with these properties is \(b = 4\), and it follows that \(2a + 1 = 19\), i.e., \(a = 9\).

Therefore, only the number 94 can satisfy the conditions of the problem. It does satisfy these conditions; indeed, \(9 - 4 = 5\) and \(2 \cdot 49 = 98 = 94 + 4\). Thus, the number 94 satisfies the conditions of the problem.