What is the likeliest 5th divisor of a randomly chosen number with at least 5 divisors?

I agree that it's 6, but I found the percentage different.

Numbers that are divisible by 6 is definitely divisible by 1,2, and 3. We only need one more factor to make 6 into 5th factor. It's either 4 or 5. But not both since if it's divisible by both 6 will be the 6th factor. So, the number that has 1, 2, 3, 4, 6,... or 1, 2, 3, 5, 6,.... as factors are numbers that are divisible by smallest common multiplication of those numbers. That means, it has to be divisible by 12 or 30 but not 60. The percentage of numbers divisible by 12 but not 60 is 1/12 X 4/5 (as one in every 5 multiplication of 12 is divisible by 60) X 100% = 6.6666%

The percentage of numbers divisible by 30 but not 60 is 1/30 X 1/2 (because every two multiplication of 30 is divisible by 60) X 100% = 1,6666%

As they don't intersect, we can just sum the percentage. The percentage of numbers with 6 as 5th factor is 8.33%

Finding the formula is hard, though. I think it's better to try it out with brute force, but it's troublesome for large numbers.

But first, to find the number with largest percentage as nth factor we need to find the number as small as possible. The reason is because the percentage of random number having m as nth factor is obviously smaller than 1/m X 100%. But it doesn't mean that the smaller the number the better, as you will see. Having certain number as nth factor may force you to include too many primes as factors and that result in having to multiply them all, resulting in small percentage. Let's try with the first few n

I think to find the number with largest percentage of random numbers having it as nth factor we can start with the number n itself. We can right away exclude numbers that are much larger than it. But this method is quite impractical

if n=1, it's obviously 1 (100%)

for n=2, it's obviously 2 (50%)

for n=3, is it 3? If it's 3 the numbers we want need to be divisible by 6 as the factors are 1,2,3,... resulting in percentage 16,66%

if it's 4 with 1,2,4,... then it needs to be divisible by 4, but not divisible by 3, resulting in percentage 1/4 X 2/3 X 100% =16.66%

we can already exclude the numbers with prime above 4, because such number will definitely need another prime to make it into 4th place factor in it's own factor or as the factor before it. even 5 multiplied with the smallest other prime, that is 2, will result in 10, causing the percentage to be less than 10%.

The remaining number is 9. But that won't work as the percentage is less than 11,1%

For n=4, if it's 4, we have 1,2,3,4,... which means the number should be divisible by 12, and no other requirement, resulting in percentage 8.33%

If it's 5, it's either 1,2,3,5,... or 1,2,4,5,... The number required needs to be divisible by 30, or 20, but not both. So it can't be the number divisible by 60. The percentage becomes 1/30 X 1/2 X 100% + 1/20X 2/3X100% = 5%

If it's 6, it can only be 1,2,3,6,... which means the required number needs to be divisible by 6, but not by 4 or 5. That results in percentage 1/6X1/2X4/5 X100% = 6.666%

that's all for now

I think the likeliest 5th divisor is 6 (for about 12% of all numbers with at least 5 divisors... :)

The fourth prime number? The (n-1)th prime number?

I'm just guessing, though.