5. In a row, all natural numbers from 1 to 100 inclusive are written in ascending order. Under each number in this row, the product of its digits is written. The same procedure is applied to the resulting row, and so on. How many odd numbers will be in the fifth row?
Answer: 19.

Solution: Note that if a number contains an even digit in its notation, then the product of the digits will be even. However, then in all subsequent rows, the product will also be even. Let's find all the products of two digits in the multiplication table that are written using only odd digits:

$$
\begin{array}{rr}
1=1 \cdot 1 ; \quad 3=1 \cdot 3 ; & 5=1 \cdot 5 ; \quad 7=1 \cdot 7 \\
9=1 \cdot 9 ; \quad 9=3 \cdot 3 ; \quad 15=3 \cdot 5 ; \quad 35=5 \cdot 7
\end{array}
$$

All other products contain even digits. Therefore, products consisting of odd digits will only be under numbers that consist of the specified digits. The list of numbers: $1,3,5,7,9,11,13,15,17,19,31,33,35,51,53,57,71,75,91$ - a total of 19 numbers.

Comment: Correct solution - 20 points. If the correct approach is followed but one extra number is written or one number is missing - 18 points. The solution is started, and there is some progress - 5 points. The solution is started, but the progress is insignificant - 1 point. The solution is incorrect or missing - 0 points.