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axezcole
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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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axezcole
M and N are two positive integers and each of them has 20 factors. If r is the total number of prime factors of M and s is the total number of prime factors of N,then what is the maximum value of r-s?

a)0
b)1
c)2
d)6
e)26

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."
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This is how I see this problem, could anybody point out where I'm wrong?

For maximum total numbers of prime factors for M while having 20 factors: 20 factors all being prime 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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josephgmat800
This is how I see this problem, could anybody point out where I'm wrong?

For maximum total numbers of prime factors for M while having 20 factors: 20 factors all being prime 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).


I see the argument you're making here and now I understand why the replies are using a different approach than mine. Thanks.
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