Consider the number 4A87B, where it is divisible by 6. Determine the highest possible value for 'A' when 'B' is at its maximum allowable value.

Since, 6 = 2 X 3 Therefore, if a number has to be divisible by '6', then it must have to be divisible by both '2' and '3' For a number two be divisible by '2', the unit digit be the number must be either '0' or an even number Therefore, largest possible value of B' = 8 For a number to be divisible by '3', the sum of the digits of the number must be divisible by '3' When 'B' = 8, sum of the digits of the number = 4 + A + 8 + 7 + 8 = (27 + A) Next number after 27 which is divisible by '3' = 36 Therefore, 'A' = 36 - 27 = 9