What is the smallest multiple of 18 of the form 2A945B, where A and B are digits?
To find the smallest multiple of 18 in the form 2A945B, we need to find the smallest values for A and B that make the entire number divisible by 18.
The sum of the digits in 2A945B is 2 + A + 9 + 4 + 5 + B = 20 + A + B. For the number to be divisible by 18, the sum of the digits must be a multiple of 9.
The smallest values for A and B that make the sum a multiple of 9 are A = 2 and B = 7.
Therefore, the smallest multiple of 18 in the form 2A945B is 229457.