During a boring class, Anna passes the time by writing down integers one below the other. Starting from a given number, in the next row she either writes the sum of the digits of the previous number or their product. Continuing this process, she noticed that every number she wrote down was odd. How many starting values of at most six digits are there for which it is true that every number written down is odd?
Solution. The text of the problem is unfortunately ambiguous. It is not clear from the text how Anna decides at each step. If she chooses multiplication and addition at will, then for example, with the five-digit number 33333, by first multiplying the digits she gets 243, and if she multiplies again, she would get an even number, but if she changes her mind and now adds the digits, the result is 9, which is suitable. Similarly, for a two-digit number, such as 35, if she always multiplies, the result will be odd, but this is no longer true for the sum.

According to one interpretation, the problem states that a starting number is suitable if both the product and the sum of its digits are always odd.

Let's examine which numbers these are. For the product of the digits to be odd, all digits must be odd. The sum of the digits will only be odd if the number has an odd number of digits. This means that the starting number cannot be 2-, 4-, or 6-digit. Therefore, it is sufficient to examine 1-, 3-, and 5-digit numbers.

There are 5 single-digit odd numbers.

For three-digit numbers, the product and sum of the digits will only be suitable if neither the product nor the sum of the digits is a two-digit number, i.e., at most 9. The suitable numbers are: 111, 113, 131, 311, 331, 313, 133, 115, 151, 511, 117, 171, 711. Thus, there are 13 such numbers.

Finally, for five-digit numbers, the sum of the digits must also be at most 9, meaning that the starting number cannot have a digit greater than 5. The possible numbers are:

- 5 ones; this is 1 possibility.
- 4 ones and 1 three; there are 5 such numbers depending on where the 3 is.
- 4 ones and 1 five; there are also 5 such numbers.
- Finally, 3 ones and 2 threes; there are 10 such numbers.

The total number of possibilities is: \(5 + 13 + 1 + 5 + 5 + 10 = 39\). Therefore, there are 39 numbers with at most six digits, such that when the digits are multiplied or added until a single-digit number is reached, the result is always odd.

According to the second interpretation, where Anna alternates between multiplication and addition, it is much more difficult to count all the possibilities.

A number cannot be continued from a starting number if the starting (or later obtained) number has both the product and the sum of its digits even, or if the number itself is even.

The enumeration of these was demonstrated by Péter Szilágyi (Debrecen, Kossuth L. Gymnasium, 11th grade). We are now sharing this:

An odd number can be a starting number if either all its digits are odd, or there are an odd number of odd digits along with even digits. In the latter case, the sequence can only be continued by adding the digits. (For example, in the case of 3257, the sum is 17, and then the product is 7.)

Among the single-digit numbers, the odd ones are suitable, and there are 5 of them. Among the two-digit numbers, all odd numbers are suitable, since either the sum or the product of the digits will definitely be odd and at most two digits. There are 45 such numbers.

Among the three-digit numbers, those are suitable which have either all three digits odd, or one odd digit and two even digits. In this case, by adding the digits, we get a two-digit odd number, which, according to the previous, is suitable. This is half of all odd three-digit numbers, i.e., 225.

Similarly, among the five-digit starting numbers, those are suitable which have 1, 3, or 5 odd digits, and the rest are even. This is half of all odd five-digit numbers, i.e., 22500.

Among the four-digit numbers, those are suitable which have 1 or 3 odd digits, and the rest are even. This is half of all odd four-digit numbers, i.e., 2250. In addition, there are 625 four-digit numbers where all digits are odd, but not all of these are suitable. Similarly, among the six-digit numbers, those are suitable which have 1, 3, or 5 odd digits, and there are 225000 such numbers; the number of six-digit numbers where all digits are odd is \(5^6 = 15625\), and not all of these are suitable. The unsuitable 4- and 6-digit numbers, considering their large number, were filtered out by the solver using a computer program. Finally, the following result was obtained:

\[
\begin{array}{rr}
1 \text { 1-digit number } & 5 \text { are } \\
2 \text { 2-digit number } & 45 \text { are } \\
3 \text { 3-digit number } & 225 \text { are } \\
\text { 4-digit number } & 2682 \text { are } \\
5 \text { 5-digit number } & 22500 \text { are } \\
6 \text { 6-digit number } & 232739 \text { are } \\
\hline \text { in total: } & 258196 .
\end{array}
\]