To solve this problem, we are looking for the probability that the product of the three results of rolling a fair six-sided die has a last digit of 0.

Step 1: Recognize when the product's last digit will be 0

The product of three numbers will have a last digit of 0 if and only if the product contains at least one factor of 5. This is because the last digit of a product is influenced by the factors of 2 and 5, and for the product to end in 0, it must include at least one factor of 5.

Step 2: Identify when the die rolls a number with a factor of 5

On a six-sided die, the possible outcomes are 1, 2, 3, 4, 5, and 6. Only the number 5 contains a factor of 5. Therefore, at least one of the three rolls must result in a 5 for the product to end in 0.

Step 3: Calculate the complement probability (when no 5 is rolled)

The complement event is the scenario where none of the three rolls results in a 5. The probability of rolling any number other than 5 is:

$P(\text{not 5}) = \frac{5}{6}$

Since the die is rolled three times, the probability that none of the three rolls results in a 5 is:

$P(\text{no 5 in three rolls}) = \left( \frac{5}{6} \right)^3 = \frac{125}{216}$

Step 4: Calculate the desired probability (at least one 5 is rolled)

The probability that at least one roll results in a 5 is the complement of the probability that none of the rolls results in a 5:

$P(\text{at least one 5}) = 1 - P(\text{no 5 in three rolls}) = 1 - \frac{125}{216} = \frac{216 - 125}{216} = \frac{91}{216}$

Step 5: Simplify the fraction

The fraction $\frac{91}{216}$ is already in its simplest form because 91 and 216 have no common factors (they are relatively prime).

Step 6: Find $m + n$

Here, $m = 91$ and $n = 216$, so:

$m + n = 91 + 216 = 307$

Thus, the answer is $\boxed{307}$.

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Do you need further clarification on any step? Here are five related questions:

1. What is the probability that the product of two dice rolls ends with a 5?
2. If the die is rolled four times, what is the probability that the product ends with a 0?
3. What is the general method for calculating probabilities involving complements?
4. How do prime factorization and number theory help in probability questions involving products?
5. Can you explain the difference between independent and dependent events in probability?

Tip: Using complement probabilities is often simpler when calculating the probability of "at least one" event occurring.