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prime number

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Source: — Problem Solving |
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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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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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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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by ArunangsuSahu » Fri Nov 11, 2011 9:03 pm
If the prime not equal to 2 then it is 2,4,2p,4p
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