prime number

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nidhis.1408: Master | Next Rank: 500 Posts; Posts: 135; Joined: Mon Oct 03, 2011 6:54 am; Followed by:4 members

prime number

by nidhis.1408 » Fri Nov 11, 2011 11:23 am

If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n?
a. two
b. three
c. four
d. six
e. eight

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Source: Beat The GMAT — Problem Solving | Fri Nov 11, 2011 11:23 am

shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 am; Thanked: 230 times; Followed by:21 members

by shankar.ashwin » Fri Nov 11, 2011 11:55 am

just substitute numbers, say p = 5

n = 4*5 = 20

even factors of 20 - 2,4,10,20 - 4 factors. C

Since its a generic problem, the pattern would be true for all primes > 2

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CappyAA: Master | Next Rank: 500 Posts; Posts: 124; Joined: Mon Jun 16, 2008 11:07 am; Thanked: 21 times; Followed by:14 members; GMAT Score:750

by CappyAA » Fri Nov 11, 2011 1:26 pm

You can also do this by factoring. n will have 3 prime factors: 2, 2, p

So the total number of even factors will be all combinations of that yields an even factor. These would be:

2
2*2 = 4
2*p = 2p
2*2*p = 4p

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user123321: Master | Next Rank: 500 Posts; Posts: 385; Joined: Fri Sep 23, 2011 9:02 pm; Thanked: 62 times; Followed by:6 members

by user123321 » Fri Nov 11, 2011 1:27 pm

nidhis.1408 wrote:If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n?
a. two
b. three
c. four
d. six
e. eight

n = 2^2 . p^1 (since p is prime, we cannot factorize it anymore)
total factors available = (2+1).(1+1) = 6
of which 1 and p are two factors which are not even divisors for n. So 6-2 = 4 factors.

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