Question
If '32x47y' is a six-digit number that is divisible by
both 11 and 12, what could be a possible value of (x + y)?Solution
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 + x + 4 + 7 + y = 16 + x + y If, we check the given options, only (x + y) = 11, makes (16 + x + y) divisible by 3 Therefore, (x + y) = 11
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