Question

Question 63

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

Answer 1

Given, 31k2 is divisible by 6. Then, it is also divisible by 2 and 3 both.
Now, 31k2 is divisible by 3, sum of its digits is a multiple of 3.
i.e. 3 + 1 + k + 2 = 0, 3, 6, 9, 12, .....
$\Rightarrow$ k + 6 = 0, 3, 6, 9, 12
$\Rightarrow$ k = 0 or 3, 6, 9