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First I chose E because I was of the opinion we need to get the count of the different prime factors and I did not know how many times the prime factors would be repeated. Then reworked to come to the opinion that "C" should be right answer.

Statement 1 : N is a factor of 7200. Could be 360, 180, 24, 3, 2, 5 etc... There could be different combination of prime factors possible. Statement 2 : 180 is a factor of N. Could be again 180, 360, 1260 etc...There could be different number of prime factors possible.

Statement 1 + Statement 2 = i)N is less than or equal to 7200 and greater than or equal to 180. ii)In order to satisfy both statements the number "N" when factorized will contain only 3 prime factors - (2^n)*(3^m)*(5^p)
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Last edited by Pansi on 06 Oct 2012, 01:58, edited 2 times in total.

(1) 7200 has plenty of divisors, 1, 2, 3, 4, 6, ... Not sufficient.

(2) There are many multiples of 180, so N can be 180, 360, 540,..., 7*180,... Not sufficient.

(1) and (2): We can deduce that 7200 = AN, for some integer A. Also, N = 180B, for some integer B. In conclusion, 7200 = 180AB, from which AB = 40. B must be a factor of 40. The prime factors of 180 are 2, 3, and 5. The prime factors of 40 are 2 and 5. In conclusion, the prime factors of N = 180B are 2, 3 and 5 only. Sufficient.

Answer C.
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(1) 7200 has plenty of divisors, 1, 2, 3, 4, 6, ... Not sufficient.

(2) There are many multiples of 180, so N can be 180, 360, 540,..., 7*180,... Not sufficient.

(1) and (2): We can deduce that 7200 = AN, for some integer A. Also, N = 180B, for some integer B. In conclusion, 7200 = 180AB, from which AB = 40. B must be a factor of 40. The prime factors of 180 are 2, 3, and 5. The prime factors of 40 are 2 and 5. In conclusion, the prime factors of N = 180B are 2, 3 and 5 only. Sufficient.

Answer C.

It just occurred to me that I could summarize the the above reasoning for (1) and (2) in the following way:

N is a divisor of 7200, which has prime factors 2, 3, and 5. Therefore, N can have at most three prime divisors: 2, 3, and 5. N must not have all of them. Since 180 is a factor of N and has prime factors 2, 3, and 5, necessarily N must have at least the same prime factors, and ALL of them. Therefore, in conclusion, N must have exactly three prime factor: 2, 3, and 5.
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Re: How many prime factors does N have? [#permalink]

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06 Oct 2012, 02:27

Zinsch123 wrote:

Wouldn't a real gmat question state that we are looking for the number of distinct prime factors?

On the GMAT yes, I think so. They are quite pedantic, so almost sure, they would explicitly stress "distinct". But, for example, there is no particular reason to count how many prime factors 32 has. You are either interested in how many (distinct) prime factors a number has, or how many distinct factors it has. I don't think you would count 5 prime factors of 32...
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Re: How many prime factors does N have? [#permalink]

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14 Jan 2014, 23:18

1

This post was BOOKMARKED

shivanigs wrote:

How many prime factors does N have?

(1) N is a factor of 7200 (2) 180 is a factor of N

Sol: As per St 1, we have 7200/N= I where I is some integer Factorizing 3600, we get : 7200= (2^5*3^2*5^2)/ I = N Now if I is 1 then N will have 2,3 and 5 as prime factors But I =25 then N will have only 2 and 3 as prime factors

Thus at the most N can have 3 prime factors but it can have less as well So A and D ruled out. This statement basically says the number of Prime factors of N will be less than or equal to 3

St2 N= 180 *A where A is some Integer N= 2^2*3^2*5 *A Thus N will surely have at 2,3 and 5 as prime factors but Integer A can take any value ie. It can be 6 or some other prime say 11 or multiple of 2 primes say 11 and 13.

So B ruled out. This statement tells us that number of prime factors of N will be greater than or equal to 3.

Combining we get we N has 3 factors only

Ans C
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Re: How many prime factors does N have? [#permalink]

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27 Sep 2015, 08:13

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Re: How many prime factors does N have? [#permalink]

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15 Jan 2017, 08:40

Great Question. Here is what i did in this one => We need to get the number of prime factors of N.

Statement 1-> N is a factor of 7200 7200=2^5*3^2*5 Hence N can have --> Zero prime factor if N=1 One prime factor Two prime factors Or Three Prime factors. Hence not sufficient.

Statement 2-> 180 is a factor of N. Hence N=2^2*3^2*5*k for some integer K. Hene N must have 2,3,5 as its prime factors. Other than that N can have any Prime factors ≥3 Hence not sufficient.

Combining the two statements => N must have exactly three prime factors> Hence sufficient.

Hence C. _________________

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Re: How many prime factors does N have?
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15 Jan 2017, 08:40

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