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

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

$31k2\phantom{\rule{0ex}{0ex}}\mathrm{sin}cetheunitdigitiseventhereforeitsatisfiesthedivisibilityrueof2\phantom{\rule{0ex}{0ex}}also,3+1+k+2=6+k\phantom{\rule{0ex}{0ex}}whenk=0,6+0=6\phantom{\rule{0ex}{0ex}}itsatisfiesthedivisibilityrueof3\phantom{\rule{0ex}{0ex}}\Rightarrow k=0$

Regards

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