## Task B-1.1.

How many ordered pairs of natural numbers $(a, b)$ are there such that $a+b=1000$ and neither of the numbers $a$ and $b$ contains the digit 0 in their decimal representation?
## First Solution.

Notice that by choosing the number $a$, the second element of the ordered pair is exactly determined, that is, $b=1000-a$. Therefore, it is sufficient to determine all possibilities for the number $a$.

There are 9 one-digit numbers that do not contain 0 in their representation, 9 \cdot 9 two-digit numbers, and 9 \cdot 9 \cdot 9 three-digit numbers. This totals 819 numbers $a$ that do not have 0 in their decimal representation.

However, among these numbers, there are also numbers $a$ that do not contain 0 in their decimal representation, but the numbers $b=1000-a$ do contain 0. These are the numbers whose tens digit is 9, that is: $91,92, \ldots, 99,191, \ldots, 199,291, \ldots, 299, \ldots 899$.

There are 9 two-digit numbers and 8 \cdot 9 three-digit numbers of such $a$, which totals 81.

Thus, the number of ordered pairs of natural numbers $(a, b)$ such that $a+b=1000$ and neither $a$ nor $b$ contains the digit 0 in their decimal representation is $819-81=738$.