A five-digit number 42A5B is divisible by 45, where A and B are single digits. What is the largest such number?

Explanation

1. 01

   $45 = 5 \times 9$, so B is 0 or 5 and the digit sum $11 + A + B$ must be a multiple of 9.

2. 02

   If $B = 0$: $11 + A$ must be 18, so $A = 7$, giving 42,750.

3. 03

   If $B = 5$: $16 + A$ must be 18, so $A = 2$, giving 42,255.

4. 04

   The larger of the two is 42,750.