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      Question

      If '38k721' is divisible by 3, then what is the highest

      possible value of 'k'?
      A 7 Correct Answer Incorrect Answer
      B 9 Correct Answer Incorrect Answer
      C 6 Correct Answer Incorrect Answer
      D 1 Correct Answer Incorrect Answer

      Solution

      ATQ,

      A number is divisible by 3 when the sum of its digits is divisible by 3.

      Sum of digits of '38k721' = 3 + 8 + k + 7 + 2 + 1 = (k + 21)

      (k + 21) is divisible by 3 when k = 0, 3, 6 or 9.

      Therefore, maximum value of 'k' = 9

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