A lock has a combination that is a four-digit positive integer. The first digit is 4 and the four-digit combination is divisible by 45 . How many different possible combinations are there?
We want to determine the number of four-digit positive integers of the form $4 x y z$ that are divisible by 45 .

A positive integer is divisible by 45 exactly when it is divisible by both 5 and 9 .

For an integer to be divisible by 5 , its units digit must be 0 or 5 .

Thus, such numbers are of the form $4 x y 0$ or $4 x y 5$.

For an integer to be divisible by 9 , the sum of its digits must be divisible by 9 .

Case 1: $4 x y 0$

The sum of the digits of $4 x y 0$ is $4+x+y$.

Since $x$ and $y$ are between 0 and 9 , inclusive, then $0 \leq x+y \leq 18$.

For $4+x+y$ to be divisible by 9 , we need $x+y=5$ or $x+y=14$.

If $x+y=5$, then $(x, y)=(0,5),(1,4),(2,3),(3,2),(4,1),(5,0)$.

If $x+y=14$, then $(x, y)=(5,9),(6,8),(7,7),(8,6),(9,5)$.

There are 11 possible four-digit numbers in this case.

Case 1: $4 x y 5$

The sum of the digits of $4 x y 5$ is $9+x+y$.

Since $x$ and $y$ are between 0 and 9 , inclusive, then $0 \leq x+y \leq 18$.

For $9+x+y$ to be divisible by 9 , we need $x+y=0$ or $x+y=9$ or $x+y=18$.

If $x+y=0$, then $(x, y)=(0,0)$.

If $x+y=9$, then $(x, y)=(0,9),(1,8),(2,7),(3,6),(4,5)$ and their reverses.

If $x+y=18$, then $(x, y)=(9,9)$.

There are 12 possible four-digit numbers in this case.

Overall, there are $11+12=23$ possible combinations.

ANSWER: 23