Two natural numbers form a pair when they have the same number of digits and their sum only contains the digit 9. For example, 225 and 774 form a pair, because both have three digits and \(225 + 774 = 999\).

a) What is the number that forms a pair with 2010?

b) How many pairs are formed by two-digit numbers?

Special pairs are pairs in which the two numbers have the same digits and, in each number, the digits are distinct. For example, 36 and 63 form a special pair, but 277 and 722 do not.

c) Give an example of a special pair with four-digit numbers.

d) Explain why there are no special pairs with three-digit numbers.
Solution

a) The number that forms a pair with 2010 is 7989, because both have 4 digits and their sum is $2010+7989=$ 9999.

b) There are ninety two-digit numbers, namely, the numbers from 10 to 99. Of these numbers, only those that start with 9 do not have a pair, that is, the ten numbers from 90 to 99. Therefore, eighty two-digit numbers have a pair to form a couple, and thus there are forty distinct couples with two digits.
c) We give the following three examples of special couples: $(2376,7623),(5814,4185)$ and $(8901,1098)$.

d) 1 $1^{\text {st }}$ solution: Let's assume there exists a special couple of three-digit numbers. Let $A$ be the hundreds digit, $B$ the tens digit, and $C$ the units digit of one of the numbers in this couple; this number is then $A B C$, noting that $A$ is not equal to 0. These are also the digits of the second number in the couple, which can then be $A B C, A C B, B A C, B C A, C A B$ or $C B A$. We then have the following six possibilities:

$$
+\begin{array}{lll}
A & B & C \\
A & B & C \\
\hline 9 & 9 & 9
\end{array}+\begin{array}{ccc}
A & B & C \\
A & C & B \\
\hline 9 & 9 & 9
\end{array}+\begin{array}{ccc}
A & B & C \\
B & A & C \\
\hline 9 & 9 & 9
\end{array}+\begin{array}{ccc}
A & B & C \\
B & C & A \\
\hline 9 & 9 & 9
\end{array}+\begin{array}{ccc}
A & B & C \\
C & A & B \\
\hline 9 & 9 & 9
\end{array}+\begin{array}{ccc}
A & B & C \\
C & B & A \\
\hline 9 & 9 & 9
\end{array}
$$

The first possibility cannot occur, as $A+A=9$ is impossible. Similarly, we eliminate the second, third, and last possibilities. In the fourth possibility, we have $B+C=9=A+C$ and it follows that $A=B$, which cannot happen, as in a special couple the digits are distinct. The same argument eliminates the fifth possibility, and thus we conclude that there are no special couples with three-digit numbers.

$2^{\text{nd}}$ solution: Suppose there exists a special couple with three-digit numbers and let $A, B$ and $C$ be the digits of these numbers. Each digit of one of the numbers, added to some digit of the second number, results in 9; thus we must have $A+A+B+B+C+C=2(A+B+C)=27$, which cannot happen, as 27 is odd. Therefore, there are no special couples with three-digit numbers.