If '32p47q' is a six-digit number that is divisible by both 11 and 12, then which of the following can be a possible value of (p + q).
Since, the given number is divisible by 12, it must also be divisible by 3 In order to de divisible by 3, the sum of digits of the number must be divisible by 3 as well. Sum of digits of the given number = 3 + 2 + p + 4 + 7 + q = 16 + p + q If, we check the given options, only (p + q) = 11, makes (16 + p + q) divisible by 3 Therefore, (p + q) = 11
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