Find the value of
3k
3
+4

​
such that the 6-digit number 310k0k is divisible by 6.

For a number to be divisible by 6, it must be simultaneously divisible by both 2 and 3.
1. Divisibility by 2:
The unit digit of the number must be even. The unit digit of 310k0k is k, so k must be an even single digit: k∈{0,2,4,6,8}.

2. Divisibility by 3:
The sum of the digits must be divisible by 3.

Sum of digits=3+1+0+k+0+k=4+2k
Let's test the even choices for k:

If k=0: Sum=4+2(0)=4 (not divisible by 3)

If k=2: Sum=4+2(2)=8 (not divisible by 3)

If k=4: Sum=4+2(4)=12 (divisible by 3) ⟹k=4 is a valid value.

If k=6: Sum=4+2(6)=16 (not divisible by 3)

If k=8: Sum=4+2(8)=20 (not divisible by 3)

Thus, k=4 is the unique solution.

Now, substitute k=4 into the target expression:

3k
3
+4

​
=
3(4)
3
+4

​
=
3(64)+4

​
=
192+4

​
=
196

​
=14