Question 42
The 7-digit number 30A210B is divisible by 9. What is the maximum possible value of (A + B)?
Solution
For a number to be divisible by 9, the sum of the digits should be divisible by 9.
Sum of digits = 6 + A + B.
The sum could be 9 or 18 or 27 or so on.
However, as A and B are single digits numbers, they both should be less than 10 and thus A + B have a maximum value of 18 in any scenario.
Case 1:
6 + A+B = 9 ==> A+B = 3.
Case 2:
6 + A+B = 18 ==> A+B = 12.
Case 3:
6 + A+B = 27 ==> A+B = 21.
Further cases are not possible because A+B exceeds 18 and thus maximum possible value of A+B for the number to be divisible by 9 is 12.
