Question

Find the value of k where 31 kz is divisible by 6​

Answer 1

Answer:

Given 31k2 is divisible by 6

For divisible by 6 it must be divisible by both 2,3

So, 31k2 will also be divisible by 2 and 3

It is divisible by 2 as there is even digit at unit place

If 31k2 is divisible by 3, sum of digit will be multiple of 3.

i.e., 3+1+k+2=0,3,6,9,12

⇒ k+6=0,3,6,9,12

But k+6=6,9,12,15 or else, k will be negative or greater than 9

⇒ k=0 or 3,6,9

Therefore , Hence proved .

I hope it may help you .