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hard prime factors

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cewalzer
 Post subject: hard prime factors
 Post Posted: Thu Mar 04, 2010 1:17 pm 
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Students


Posts: 1
How many different prime numbers are factors of the positive integer n?

(1) Four different prime numbers are factors of 2n
(2) Four different prime numbers are factor of n^2

I tried pluggin numbers....

(1) if 2n=2.3.5.7 then n= 3.5.7 (3 prime factors) , but 2n could be 2n=3.5.7.(2^2) so n= 2.3.5.7 (4 prime factors)

Not sufficient

(2) if n^2=(2^2).(3^2).(5^2).(7^2) then n=2.3.5.7 =>sufficient

So answer is B.... but, is there any theoretical way to solve this?
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cyapt81
 Post subject: Re: hard prime factors
 Post Posted: Sat Mar 20, 2010 4:24 pm 
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(1) If 2n has four different prime #'s that means the prime number 2 must be a factor of 2n. If n itself has three distinct prime factors, e.g. 3, 5 and 7, 2n would have 4 factors: 2, 3, 4, 7. If on the other hand n has 4 distinct prime factors one of which is 2, e.g. 2, 3, 5, 7 then n has four distinct prime factors and 2n still has 4 distinct prime factors, 2, 3, 5, and 7. So n could have 3 or 4 distinct prime factors. INSUFFICIENT.

(2) Since n^2 is made up of two chains of n's, the only way n^2 will have 4 distinct prime factors is if n itself has 4 distinct prime factors e.g. if n has the prime factors 2, 3, 5 and 7, n^2 will have only those prime factors as well since two chains of n, (2x3x5x7)(2x3x5x7) will equal n^2. SUFFICIENT

B.
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mschwrtz
 Post subject: Re: hard prime factors
 Post Posted: Tue Mar 30, 2010 4:35 pm 
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ManhattanGMAT Staff


Posts: 506
Nice explanation cyapt81. Thanks.

By the way, cewalzer, I wonder if you were sensitive to the same point. Was it just a coincidence that you tried one number with a factor of 2 and a second without a factor of 2?

I suspect that you saw that even values for n might given you a different answer than odd value, and picked values accordingly. If so, good on you. That's just what you want to do when testing values in DS: ask what values are likely to yield different results.
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