A number with 20 factors can be written as
20 = 2*2*5 or\(a^1 * b^1 * c^4\)
20 = 4*5 or\(a^3 * b^4\)
20 = 10*2 or \(a^9 * b^1\)
20 = 20 or \(a^19\)

where a, b, c are prime factors
Hence a number with 20 factors can have a maximum of 3 and a minimum of 1 prime factors
3-1 = 2
Answer C
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For the maximum value of r - s, let's find the maximum value of r and the minimum value of s.

Given that r is the total number of prime factors of M, and M is a positive integer that has 20 factors.
For the maximum value of r, let's assume that M = prime number \(p1^{1}p2^{1}p3^{4}\). In such a case, the number of factors of M is 20 and the number of prime factors is 3. We cannot have more than 3 prime numbers for the number of factors to be 20.
Hence, the lowest possible value of r is 3.

Also, Given that s is the total number of prime factors of N, and N is a positive integer that has 20 factors.
For the minimum value of s, let's assume that N = prime number \(p^{19}\). In such a case, the number of factors of N is 20 and the number of prime factors is 1.
Hence, the lowest possible value of s is 1.

Therefore, Max (r) - Min (s) = 3 - 1 = 2. So, the correct answer is option C. 2
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Hi axezcole92,

This question should be in the PS subforum. Please read here:
https://gmatclub.com/forum/rules-for-po ... 33935.html
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This is how I see this problem, could anybody point out where I'm wrong?

If M and N are both integers that have 20 factors, we could hypothetically assume:
(P for prime numbers)

For maximum total numbers of prime factors for M while having 20 factors: 20 factors all being prime factors (which we would like M to be if we would like to maximize r, and thus maximizing r-s)
M=P1*P2*...*P20, therefore r=20

For minimum total numbers of prime factors for N while having 20 factors: 20 factors all being the same prime factor (which we would like N to be if we would like to minimize s, and thus maximizing r-s)
N=P1^20, therefore s=1
r-s=19(#)

Why isn't there a 19 in the options?

Am I missing anything because I don't understand why other replies stick to "the factor of 20" instead of "they have 20 factors."

If a number has 20 different prime factors, then the number will have a lot more than 20 factors in total. In fact, such a number would have at least 2^20 divisors in total, so would have more than a million divisors.

You could look at a simpler example - say a number has three prime divisors, so perhaps the number is (2)(3)(5) = 30. That number has eight divisors in total: 1, 2, 3, 5, 6, 10, 15, and 30. Because any combination of our prime divisors is also a divisor of 30, we have several divisors that are not prime. So that's the reason the number in this question cannot have nearly as many as twenty different prime factors.

There's a method you can learn that lets you count any number's divisors once you have that number's prime factorization. The solutions above are all using that method. If you haven't encountered that before, then I wouldn't start with the question posted here, which tests the concept in a much harder way than most questions do. I'd suggest you learn the theory first, from a good prep book, and then look at some easier questions before trying this one (and you could probably even skip this one, because I don't think you'll see something like it on the GMAT).
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I see the argument you're making here and now I understand why the replies are using a different approach than mine. Thanks.