K^4 is an integer with unit digit of 6. K is also an integer. How many

\(K^4\) is an integer with unit digit of 6 and K is also an integer
Now, we need to find how many different integers are possible for the unit digit of integer \(K\)

Let's understand the problem with one theory
To find unit digit of power of a number we only need to know the units digits of the number
Ex: \(12^2\) = 144
Now, units digit of 144 is 4 and to find this we only need to know the units digit of 12 which is 2. [As \(2^2\) = 4]

So, to find units digit of \(K^4\) we only need to know the units digit of K
Now, Units digit of \(K^4\) = 6
So, we need to find out numbers whose \((units digit)^4\) will give us 6
We need to consider numbers from 0 to 9. Not we need 6 is the units digit of \(K^4\) so that means that K has to even number.
So, we can ignore 1,3,5,7,9 and should consider only 2,4,6,8,0 as the possible units digit for K
Units digit of \(0^4\) will be 0 => NOT possible ( \(0^4\) = 0)
Units digit of \(2^4\) will be 6 => Possible ( \(2^4\) = 16 )
Units digit of \(4^4\) will be 6 => Possible ( \(4^4\) = \(2^8\) = 256 )
Units digit of \(6^4\) will be 6 => Possible ( Units digit of all positive powers of 6 is 6 )
Units digit of \(8^4\) will be 6 => Possible ( \(8^2\) = 64, now \(8^4\) can we written as \(8^2\) * \(8^2\) = 64 * 64 => Multiply units digit so 4*4 will give us 6 as the units digit of \(8^4\) )

So, answer will be 2,4,6,8
So, Answer will be D

Hope it helps!