32a6b is divisible by 18. What can be the maximum value of a+b?Options8161825

The given number is 32a6b. We know that a number is divisible by 18 if it is divisible by both 2 and 9.

1. Divisibility by 2: A number is divisible by 2 if its last digit is 0, 2, 4, 6, or 8. Here, the last digit is 'b', so for divisibility by 2, b can be 0, 2, 4, 6, or 8.

2. Divisibility by 9: A number is divisible by 9 if the sum of its digits is a multiple of 9. Here, the sum of the digits is 3+2+a+6+b = 11+a+b. For this to be a multiple of 9, a+b can be -2, 7, 16, 25, 34, etc. But since a and b are digits, they must be between 0 and 9. So, the possible values of a+b are -2 (which is not possible as a and b are positive) and 7.

Therefore, the maximum value of a+b is 7. So, the correct option is not listed in the given options.