A number was written on the board. At each step, we add to it the largest of its digits (for example, if we have written 142, the next number will be 146). What is the maximum number of odd numbers that can be written consecutively by proceeding in this way?

## High school statements
The answer is 5. Suppose we start from an odd number $n$. We denote $n_{i}$ as the $i$-th number written, with $n_{1}=n$. Let $c_{i}$ and $d_{i}$ be the largest digit and the units digit of $n_{i}$, respectively. If $c_{1}$ is odd, then $n_{2}=n_{1}+c_{1}$ is even, and we have written only one odd number. Note also that $c_{1}$ cannot be equal to 0.

If $c_{1}=2$, then $d_{1}=1$. We have $n_{2}=n_{1}+2$, so $c_{2}=3$ and $d_{2}=3$, and $n_{3}=n_{2}+3$ is even. The third number written is therefore even.

If $c_{1}=4$, then $d_{1}$ can be 1 or 3. If $d_{1}=1$, then $n_{2}=n_{1}+4$ so $c_{2}=5$ and $d_{2}=5$, and $n_{3}=n_{2}+5$ is even. If $d_{1}=3$, then $n_{2}=n_{1}+4$ so $c_{2}=7$ and $d_{2}=7$, and $n_{3}=n_{2}+7$ is even.

If $c_{1}=6$, then $d_{1}$ can be 1, 3, or 5. In the first two cases, we get $n_{3}$ even as before. If $d_{1}=5$, then $n_{2}=n_{1}+6$ so $d_{2}=1$. Moreover, the largest digit either remains the same ($c_{2}=6$) or increases by 1 ($c_{2}=7$). In the latter case, $n_{3}=n_{2}+7$ is even. In the former, $n_{3}=n_{2}+6$ so $d_{3}=7$ and $c_{3}=7$ so $n_{4}=n_{3}+7$ is even.

If $c_{1}=8$, then $d_{1}$ can be 1, 3, 5, or 7.

$\nabla$ if $d_{1}=1$, we get $c_{2}=d_{2}=9$ so $n_{3}$ is even.

$\triangleright$ if $d_{1}=3$, then $d_{2}=1$ and $c_{2}$ can be 8 or 9. In the latter case, $n_{3}$ is even, and in the former, we are reduced to the previous case ($c=8, d=1$) so $n_{4}$ is even.

$\triangleright$ if $d_{1}=5$, then $d_{2}=3$ and $c_{2}$ can be 8 or 9. In the latter case, $n_{3}$ is even, and in the former, we are reduced to the previous case ($c=8, d=3$) so $n_{4}$ or $n_{5}$ is even.

$\triangleright$ if $d_{1}=7$, then $d_{2}=5$ and $c_{2}$ can be 8 or 9. In the latter case, $n_{3}$ is even, and in the former, we are reduced to the previous case ($c=8, d=5$) so $n_{4}$ or $n_{5}$ or $n_{6}$ is even.

It is therefore impossible to write 6 consecutive odd integers. However, if we start with 807, we will write successively $807, 815, 823, 831$, and 839, so we can write 5 consecutive odd integers.