Complete 54?? to Make It Divisible by 6: Missing Digit Challenge

Complete the number so that it is divisible by 6 without a remainder:

$54??$

To solve this problem, we need to find digits to replace the question marks in $54??$ such that the resulting number is divisible by 6. We'll use the divisibility rules for both 2 and 3.

First, since the number must be even, the last digit must be one of the even numbers: 0, 2, 4, 6, or 8.

Second, we need the sum of the digits to be divisible by 3. The initial digits '54' have a sum of $5 + 4 = 9$. Since 9 is already divisible by 3, we need the sum of the missing digits (let’s denote them as x and y) to also result in a number divisible by 3 when added to 9.

Therefore, we need $9 + x + y \equiv 0 \pmod{3}$. This simplifies to $x + y \equiv 0 \pmod{3}$.

Let us check with the provided options:

• Option 1: $3,3$: The sum $x + y = 3 + 3 = 6$. Divisible by 3. Last digit is 3 (odd), this does not satisfy divisibility by 2.
• Option 2: $6,8$: The sum $x + y = 6 + 8 = 14$. Not divisible by 3.
• Option 3: $5,5$: The sum $x + y = 5 + 5 = 10$. Not divisible by 3.
• Option 4: $3,1$: The sum $x + y = 3 + 1 = 4$. Not divisible by 3.

Since none of the provided options satisfy both divisibility conditions, the correct answer is "None of the above".

In conclusion, Option 5: "None of the above" is the correct choice.