Which of the following can be the value of ‘k’ so that the number 217924k is divisible by 6?

To find the value of $k$ such that the number $217924k$ is divisible by 6, we must apply the rules of divisibility for 6.

A number is divisible by 6 if and only if it is divisible by both 2 and 3.

Step 1: Check divisibility by 2
For a number to be divisible by 2, its last digit must be an even number (0, 2, 4, 6, or 8).
Here, the last digit is $k$. Since all the given options (2, 6, 0, and 4) are even numbers, the divisibility rule for 2 is satisfied for all of them.

Step 2: Check divisibility by 3
For a number to be divisible by 3, the sum of its digits must be divisible by 3.
Let us calculate the sum of the digits of the number $217924k$:

$\text{Sum of digits}=2+1+7+9+2+4+k$

$\text{Sum of digits}=25+k$

Now, we substitute each of the given options for $k$ to see which one results in a sum that is a multiple of 3:

- If $k=2$:
The sum is $25+2=27$, which is divisible by 3 (since $27/3=9$).

- If $k=6$:
The sum is $25+6=31$, which is not divisible by 3.

- If $k=0$:
The sum is $25+0=25$, which is not divisible by 3.

- If $k=4$:
The sum is $25+4=29$, which is not divisible by 3.

Thus, the only value of $k$ from the choices that makes the number divisible by 6 is 2.