Question 705005

Let the number be $d_1 d_2 d_3$. We require $d_1 + d_3 = 9$. Since $d_1$ is the first digit, $d_1 \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. Pairs $(d_1, d_3)$ are: $(1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1), (9,0)$. Total 9 pairs. For each pair, the middle digit $d_2$ can be any value from 0 to 9 (10 possibilities). Total numbers = $9 \times 10 = 90$.