Exuberant Numbers

Let the number be $10a + b$ , where $1 \leq a \leq 9$ . Fix $b$ , then we want $10a$ to be a multiple of $b$ .

Case 1: $b=1$ . Then $10a$ is a multiple of 1, so $a$ is a multiple of 1 - 9 possibilities.

Case 2: $b=2$ . Then $10a$ is a multiple of 2, so $a$ is a multiple of 1 - 9 possibilities.

Case 3: $b=3$ . Then $10a$ is a multiple of 3, so $a$ is a multiple of 3 - 3 possibilities.

Case 4: $b=4$ . Then $10a$ is a multiple of 4, so $a$ is a multiple of 2 - 4 possibilities.

Case 5: $b=5$ . Then $10a$ is a multiple of 5, so $a$ is a multiple of 1 - 9 possibilities.

Case 6: $b=6$ . Then $10a$ is a multiple of 6, so $a$ is a multiple of 3 - 3 possibilities.

Case 7: $b=7$ . Then $10a$ is a multiple of 7, so $a$ is a multiple of 7 - 1 possibilities.

Case 8: $b=8$ . Then $10a$ is a multiple of 8, so $a$ is a multiple of 4 - 2 possibilities.

Case 9: $b=9$ . Then $10a$ is a multiple of 9, so $a$ is a multiple of 9 - 1 possibilities.

Case 9: $b=0$ . Then $10a$ is a multiple of 0, so $a$ is a multiple of 0 - 0 possibilities.

By the rule of sum, there are $9 + 9+3+4+9+3+1+2+1 + 0= 41$ such numbers.