Ok so I think you misunderstood my statement, and I did submit my answer. I SUBMITTED MY ANSWER.
"You correctly used constructive counting within casework to count the number of factorific colorings and total number of colorings, then used these values to calculate the desired probability.
Your casework to count the successes did work, but isn't the most efficient such approach as you noted. Plus, it is hard to check some of your cases to ensure you found all of the successful colorings and did not miss any, although your method of "combining" smaller sets could have helped you do this. Can you think of a faster and simpler approach here? You may still need some form of casework, although not necessarily based on the number of blue numbers. There are other approaches that have fewer cases and less complex ways to count the successes within each case.
Your second case when counting the successes was not labeled correctly. Also, it is a bit confusing to label the cases with bolded combinations, since some of those technically have the same value. Instead, you could mark them with a short description like "Exactly numbers colored:" and so on."
This comment was from a reviewer on AoPS
And I did not submit your answer, I submitted mine.
And If you do want to see the answer OFFICIALLY posted by AoPS, here it is
We will compute this asThe total number of colorings is just because there are two choices for each number, it can be either blue or not blue.
The number of factorific colorings is a bit more difficult to compute. First, we notice that every factorific coloring has 1 colored blue.
So, we only have to deal with the numbers 2 through 6. The only overlap in divisors comes from 2 being a factor of both 4 and 6, so we count by casework on whether 4 and 6 are colored blue.
Case 1: 6 is colored blue and 4 is not.
In this case, we must color 2 and 3 blue, because they are divisors of 6.
We know 4 is not colored blue, and 5 can go either way. So, there are 2 such colorings.
Case 2: 6 and 4 are both blue.
In this case, we must color 2 and 3 blue, because they are divisors of 6.Again, 5 can go either way. So, there are again 2 colorings in this case.
Case 3: 4 is blue and 6 is not.
In this case, we know that 2 is colored blue and 6 is not.There are two options for both 3 and 5. So there are such colorings in this case.
Case 4: Neither 4 nor 6 is blue.
In this case, we can simply color 2, 3, and 5 however we want. There are ways to do that.
Combining our work, we find that there are factorific colorings. So, the probability that Grogg's coloring is factorific is 1/4.
1/4. YES MY ANSWER IS CORRECT.
And it is not nice to go around media like this with an attitude like that.
