jollyjellyjojo

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. 

I guess I submitted my answer and got it right???? It is a bit wordy 

although GingerAle I used the fact how you used your casework and how you organized them. 

And there's no need to be rude? Chill? pls? 

Uh, This work did actually belong to me, although like I said, is wordy and not the most efficient way, :/

"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."
 That was from AoPS???

"My answer is spot on.  I can “prove” my answer is correct,"

-???

"I think I need powerful anti-nausea meds to wade through this irrelevant, wordy, neurotic swill. Jesus Christ! It’s amazing that someone far surpassed Mr. BB in pasting mathematically infused slop on this forum.  It’s interesting that someone can write reasonably tight LaTex code but doesn’t know how to implement its execution here; "

-Uh...Ok??? I'm sorry if it is not pleasing to you, but  ¯\_(ツ)_/¯ 

Ok, so I'll use GingerAle's percception of the question, although I did get a different answer 

MIRB16 I also think you are in my AoPS class???

[b]The numbers $1,2,3,4,5,6$ are written down and some of them are colored blue. A coloring is called factorific if there is at least one blue number, and for each blue number, all of its divisors are also blue.

If Grogg randomly colors some, all, or none of the numbers from 1 to 6 blue, what is the probability that his coloring is factorific?[/b]

So, to find the probability, we have to first find the total number of possibilities.

If Grogg colors none of them, that is $\binom{6}{0}$ ways he can color

If Grogg colors 1 number, that is $\binom{6}{1}$ ways he can color

If Grogg colors 2 numbers, that is $\binom{6}{2}$ ways he can color

If Grogg colors 3 numbers, that is $\binom{6}{3}$ ways he can color

If Grogg colors 4 numbers, that is $\binom{6}{4}$ ways he can color

If Grogg colors 5 numbers, that is $\binom{6}{5}$ ways he can color

If Grogg colors 6 numbers, that is $\binom{6}{6}$ ways he can color

$$\binom{6}{0} + \binom{6}{1} + \binom{6}{2} + \binom{6}{3} + \binom{6}{4} + \binom{6}{5} + \binom{6}{6}$$

$$1 + 6 + 15 + 20 + 15 + 6 + 1$$

$$64$$

There are 64 ways to color numbers.

Now we have to set some rules for our casework.

For this casework, let's include the factorifics that have more than one main "blue" number, so 5 and 6 could be colored as main numbers (just an example).

PS: I will also can the numbers colored "sets"

$\bf\binom{6}{0}$

For this option, this cant work because 0 has no divisors! (You cannot divide by 0)

c
Since we're focusing on numbers with only 1 divisor, that would be ... only 1. The reason is because if you only have to have 1 divisor, then the divisor HAS to be the number itself - and hat only applies for one because when you divide 1 by 1, it is 1.

$\bf\binom{6}{2}$

Theses numbers can only have 2 divisors, and that means it can't be composite numbers, because the minimum amount of divisors a composite number can have are 3 divisors (Perfect Square Numbers). That means that the numbers are all [b]prime[/b]. The only numbers that would work are prime numbers less than 6, and those are 2, 3, and 5. So the numbers colored are
$1,2$
$1,3$
$1,5$

$\bf\binom{6}{3}$

Now here is where it gets a little bit confusing, but it is manageable, same for the rest.
So, we can first list out the perfect square numbers, (they have 3 divisors, remember) and the only one less than 6 is 4.

$1, 2, 4$

Now, the reason why the rest are a bit more confusing is because we are combining sets from the previous case to make our present case, so that means cases like, say, $1, 2, 3$ will work because $1, 3$ is one and $1, 2$ is the next. The next set of colored numbers that work are $1, 2, 5$ because again, they include two sets from the previous. The next set is $1, 3, 5$  because there are two different cases (again) 3, and 5. And if we try the next set... we see that there are no more sets left that meet the specifications.

$1, 2, 3$
$1, 2, 4$
$1, 2, 5$
$1, 3, 5$

$\bf\binom{6}{4}$

Again, it's a bit more complicated but we can handle it 

Ok, so the actual composition numbers besides perfect square numbers under or equal to 6 are only...6. 6 has the divisors, or factors, $1, 2, 3, 6$

$1, 2, 3, 6$

The next ones we will combine from the previous case.
The first set should be $1, 2, 3, 5$ because all of the sets apply here, too. The next set should be $1, 2, 3, 4$ because that applies too - the perfect square 4 and the  prime number 3 to provide this set. and lastly, we have the set $1, 2, 4, 5$ and we can apply the same concept from the set $1, 2, 3, 4$, 5 is also a prime number.

$1, 2, 3, 4$ 
$1, 2, 3, 5$
$1, 2, 3, 6$
$1, 2, 4, 5$

$\bf\binom{6}{5}$

Hang in there, we're almost done!

So, this, we clearly don't have a number with 5 divisors that is below or equal to 6, because 6 had 4 divisors and it was the largest number possible, so now all we can do is to combine sets (again) 

The first set we can try is by combining the two sets $1, 2, 3, 5$ and the set $1, 2, 4$, and what results is $1, 2, 3, 4, 5$, which works! The next possible set can be $1, 2, 3, 4, 6$, but this set is composed of the sets $1, 2, 3, 6$ and $1, 2, 4$. The last, one we can try is $1, 2, 3, 5, 6$ composed of the sets $1, 2, 3, 5$ and the set $1, 2, 3, 6$.

$1,2,3,4,5$
$1,2,3,4,6$
$1,2,3,5,6$

$\bf\binom{6}{6}$

And finally, the last possibility where Grogg colors all of the numbers. How delightful :) 

This case has only one set $1, 2, 3, 4, 5, 6$, and all we have to do is to check this if it works. we can obviously do this by combining the two sets from before $1,2,3,4,5$  and $1,2,3,5,6$. It does work! Now we can put everything together.

$\bf\binom{6}{0}$ has 0 cases

$\bf\binom{6}{1}$ has 1 case

$\bf\binom{6}{2}$ has 3 cases

$\bf\binom{6}{3}$ has 4 cases

$\bf\binom{6}{4}$ has 4 cases

$\bf\binom{6}{5}$ has 3 cases

$\bf\binom{6}{6}$ has 1 case

$\frac{1+3+4+4+3+1}{64} = \frac{16}{64} = \frac{1}{4}$ 

The probability that Grogg has colored in a factorfic is $\bf\frac{1}{4}$.

Hopefully that helped?

It is 0.16 or 4/25 yay