Complete the Number 54??: Finding the Digit for Divisibility by 6

To solve this problem, let's begin by considering the divisibility rules for 6:

• For a number to be divisible by 6, it must be divisible by both 2 and 3.

First, we analyze divisibility by 2:

• The last digit of the number (represented by '?') must be even. Possible even digits are 0, 2, 4, 6, 8.

Next, consider the divisibility by 3:

• The sum of all the digits, $5 + 4 + x + y$, must be divisible by 3.
• Given that $5 + 4 = 9$, we seek values for $x + y$ such that $9 + x + y$ is divisible by 3.

Now, let’s test possible values for $x$ (tens place) and $y$ (units place) using the conditions above:

• If $y = 0$, $x + 0$ must satisfy $9 + x + 0$ is divisible by 3 (i.e., $x + 9 \equiv 0 \mod 3$). Testing values, there’s no $x$ value making it divisible by 6.
• If $y = 2$, $x + 2$ must satisfy $9 + x + 2 \equiv 0 \mod 3$. It works when $x = 4$ because $9 + 4 + 2 = 15$ is divisible by 3.
• If $y = 4$, $x + 4$ must satisfy $9 + x + 4 \equiv 0 \mod 3$. No values of $x$ make this divisible by 6.

Therefore, the combination that satisfies both divisibility rules is $x = 4$ and $y = 2$.

Conclusion: The missing numbers making 54?? divisible by 6 are $4$ and $2$, matching the selection $4,2$ from the choices.