Question 1210180
Let the three-digit number be $abc$, where $a$, $b$, and $c$ are digits. We are given the following conditions:

1.  **Sum of digits is 12:** $a + b + c = 12$
2.  **Product of digits is 48:** $a \cdot b \cdot c = 48$
3.  **The number is divisible by 6:** This means the number is divisible by both 2 and 3.

From condition 3, for a number to be divisible by 2, the last digit $c$ must be even. For a number to be divisible by 3, the sum of its digits must be divisible by 3. Since $a + b + c = 12$, and 12 is divisible by 3, this condition is already satisfied.

Now, we need to find three digits whose sum is 12 and product is 48.
Let's find the possible sets of three digits whose product is 48.
The prime factorization of 48 is $2^4 \cdot 3$.

Possible sets of three digits:
* $1 \cdot 6 \cdot 8$
* $2 \cdot 4 \cdot 6$
* $3 \cdot 4 \cdot 4$
* $2 \cdot 3 \cdot 8$

Now, let's check which sets also satisfy the sum condition:
* $1 + 6 + 8 = 15$ (doesn't equal 12)
* $2 + 4 + 6 = 12$ (satisfies)
* $3 + 4 + 4 = 11$ (doesn't equal 12)
* $2 + 3 + 8 = 13$ (does not equal 12)

So the only set of digits that works is 2, 4, 6.

Now, we need to find the arrangements of 2, 4, 6 such that the number is divisible by 6 (i.e., the last digit is even).
Possible numbers:
* 246
* 264
* 426
* 462
* 624
* 642

All of these numbers satisfy the conditions:
* $2 + 4 + 6 = 12$
* $2 \cdot 4 \cdot 6 = 48$
* All numbers are even, so they are divisible by 2.
* All numbers have a digit sum of 12, so they are divisible by 3.
* Therefore, all numbers are divisible by 6.

So the possible numbers are 246, 264, 426, 462, 624, 642.