To determine if the seven-digit number 62AB427 is divisible by 99, we can use the divisibility rule for 99, which states that a number is divisible by 99 if the sum of its digits is divisible by 9.
So, the sum of the digits in the given number is:
6 + 2 + A + B + 4 + 2 + 7 = 21 + A + B
For the number to be divisible by 99, 21 + A + B must be divisible by 9.
Now, the possible values for A and B range from 0 to 9. We need to find the values that make 21 + A + B divisible by 9.
For 21 + A + B to be divisible by 9, the sum of A and B must be 6 (since 21 + 6 = 27, which is divisible by 9).
So, A + B = 6.
Given that the digits can range from 0 to 9, and A + B = 6, the possible pairs of A and B are (0, 6), (1, 5), (2, 4), and (3, 3).
Since A and B are digits in the number, they cannot be greater than 9. So, the only possible pair is (3, 3).
Therefore, the sum of A and B is 3 + 3 = 6.
