Find the value of k where 31k2 is divisible by 6

Dear student,
Applying the divisibility of 6 i.e a number should pass the test of 2 and 3
i.e it must be an even number and also the sum of digit is divisible by 3
31k2 since the unit digit is even therefore it satisfies the divisibility rue of 2also, 3+1+k+2=6+kwhen k=0, 6+0=6 it satisfies the divisibility rue of 3k=0


  • 6
Is the answer 0 ?
  • 2
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