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# If the square root of p2 is an integer, which of the

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If the square root of p2 is an integer, which of the [#permalink]  13 Feb 2006, 10:49
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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

I
II
III
I and II
II and III
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I and II

p2 is the square of an integer.
The square of any number can be written as a product of even prime factors(3*3, 5*5, 2*2*2*2 etc)
and has odd factors(because 1 is also a factor)
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In (III) is P the square root of P2?

(I) could be true because every number can be expressed as a product of an odd number of factors, (including 1) But Im not sure if 1 can be included or not-

e.g. : F(9) = 1*3*3 = 1*3*3*1 (both odd and even number of factors).
This one is not concrete, Ill say "No" for now.

(III) is not always true. Same argument as I. I and II are complimentary. So at this point I would just assume that, 1 is NOT to be considered as a factor. This assumption actually helps in option II

(II) is valid since P2 is a perfect square, each prime factor is repeated. (Ignore the 1 as a factor)

Hence I would opt for II only. Will be interesting to see what the OA on this is.
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Re: PS-Factors [#permalink]  14 Feb 2006, 01:12
cool_jonny009 wrote:
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

I
II
III
I and II
II and III

got II ONLY.

For I, if p^2=4, then factors are 1,2,4 (odd number of factors)
p^2=36, then 1,2,3,4,6,9,12,36 (EVEN number)

II holds.

For III if P=2 then (1,2) -- EVEN
P=4 then (1,2,4)-- ODD

Hence II
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I and II.

I. p^2 is a perfect square...will have an odd number of total factors.
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