$\underline{\text { Markelov S.V. }}$

For each of the numbers $1, 19, 199, 1999$, etc., a separate card was made and the number was written on it.

a) Is it possible to choose at least three cards such that the sum of the numbers on them equals a number, all of whose digits, except one, are twos?
b) Suppose several cards were chosen such that the sum of the numbers on them equals a number, all of whose digits, except one, are twos. What can the digit, different from two, be?
a) For example, $19+199+1999+\ldots+199999999=222222212$.

b) Example. Example with zero: $1+19=20$; example with 1 is given in part a).

![](https://cdn.mathpix.com/cropped/2024_05_06_0918ecac094e52bfe73bg-09.jpg?height=54&width=1904&top_left_y=316&top_left_x=0)
In the selected \$k cards, the sum \$S satisfies the inequality $\$ \mathrm{E} \backslash$ \leqslant $\mathrm{S}<2 \mathrm{E} \$$, where the number $\$ \mathrm{E} \$$ consists of $\$ \mathrm{k} \$$ ones and possibly several zeros. This means that the number of digits in \$S $\$$ and $2 \mathrm{E} \$$ is the same, and at least one digit of the number $\$ S \$$ is less than the corresponding digit of the number $\$ 2 E \$$, i.e., less than 2. This can only be the digit 0 or 1.

## Answer

a) yes.

b) 0 or 1.