How can we arrange five two-digit numbers, using the digits $0,1, \ldots, 9$ (each used once), so that the product of the resulting numbers is as large as possible?
To ensure that the product of the five assembled numbers is as large as possible, it is necessary that in each number, the digit in the tens place is greater than the digit in the units place. If this were not the case, the product of the numbers would certainly increase by "reversing" the number.

Thus, among the ten digits, the largest, 9, must be in the tens place, and the smallest, 0, must be in the units place. We will show that these two digits must appear in the same number. Indeed, if 9 appears with the digit \(a\) and 0 with the digit \(b\), then the product of these two numbers would be smaller if we were to swap the 0 and the \(a\):

\[
(10 \cdot 9 + a)(10 \cdot b + 0) = 900b + 10ab < 900b + 10 \cdot a \cdot 9 = (10 \cdot 9 + 0)(10b + a)
\]

since \(b < 9\). Thus, one of the two-digit numbers must certainly be 90.

Similarly, the next largest and smallest digits, 8 and 1, must also appear in the same number, because if they were to appear with the digits \(c\) and \(d\) respectively, then swapping the 1 and \(c\) would again increase the product:

\[
(10 \cdot 8 + c)(10 \cdot d + 1) = 810d + 80 + c + 10(c - 1)d < 810d + 80 + c + 10(c - 1) \cdot 8 = (10 \cdot 8 + 1)(10 \cdot d + c)
\]

Similarly, we find that the digits 7 and 2, 6 and 3, and 5 and 4 must appear in the same numbers. Thus, the five two-digit numbers we are looking for are 90, 81, 72, 63, and 54, and their product is 1785641760.

Note. Among the submitted solutions, only a very few were flawless. Most of the solvers stated without justification that the digits 5, 6, ..., 9 must appear in the tens place, although proving this is not so simple. There was a solver who argued that "there is logic in mathematics, and the only logical arrangement is the above." This "solution" naturally fell into the category of incorrect ones.