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by coolvishu11 » Sun May 10, 2009 4:24 pm
4. If the square root of p2 is an integer, which of the following must
be true?
I. p2 has an odd number of factors
II. p2 can be expressed as the product of an even number of prime
factors
III. p has an even number of factors

a· I
b· II
c· III
d· I and II
·e II and III


OA is D
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by DeepakR » Sun May 10, 2009 5:04 pm
Let p^2 = 16.

1. Factors of 16 = 1, 2, 4, 8, 16 = 5 numbers which are factors and hence #1 satisfies.

2. Prime factors of 16 = 2*2*2*2 which is the product of even number of prime factors of 16.

3. Does not satisfy due to #1

You can also check with number like 25

1. Factors of 25 = 1, 5, 25 = satisfies
2. Prime factors of 25 = 5 * 5 = we have 2 fives which are prime and even numbers (i.e two 5's)
3. Does not satisfy due to #1

Hence D.)

-Deepak
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by sureshbala » Sun May 10, 2009 7:54 pm
Since p^2 is a perfect square and every perfect square will always have odd number of factors, I is always true.

Also, since p^2 contains odd of factors, when p^2 is written in the form of prime factorization, each and every prime factor will be raised to an even power and hence p^2 can always be written as even number of prime numbers. Hence, II is true.

I contradicts III

Thus I and II are true.
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by Ian Stewart » Mon May 11, 2009 3:56 am
sureshbala wrote: I contradicts III
I think you may have misread III; III says that p has an even number of factors, not that p^2 has an even number of factors, so I and III are not contradictory.

The only numbers which have an odd number of positive divisors are perfect squares. Since p^2 is a perfect square, I must be true. III may be true, and it may not; if p is also a perfect square (e.g. if p = 4, and p^2 = 16, or p = 9 and p^2 = 81), then p will have an odd number of divisors. If p is not a perfect square, then p will have an even number of divisors.
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