2006 AIME II 第 7 题

Count the bad pairs instead. If $a$ ends in $0,$ then so does $b,$ giving $a = 10r,$ $b = 10s$ with $r + s = 100.$

If neither number ends in $0,$ a zero digit can only be a middle digit $h0u$ — and then the other number’s tens digit is $9.$

There are $999$ pairs in all ($a = 1, \ldots, 999$); count the forbidden ones. If $a$ has units digit $0,$ so does $b,$ and writing $a = 10r,$ $b = 10s$ gives $r + s = 100$ with $1 \le r \le 99:$ that is $99$ forbidden pairs.

Now suppose both units digits are nonzero. Then a number in the pair has a zero digit exactly when it is a three-digit number of the form $h0u$ with $h, u \in \{1, \ldots, 9\}$ (a one- or two-digit number with nonzero units digit has no zero digit). If $a = h0u,$ then $b = 1000 - a$ $= 100(9 - h) + 90$ $+ (10 - u)$ has tens digit $9,$ so $b$ is not also of that form. Hence the forbidden pairs here are those where exactly one of $a, b$ equals $h0u:$ $81 + 81 = 162$ pairs.

The total number of forbidden pairs is $99 + 162 = 261,$ so the answer is $999 - 261 = 738.$