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# prime numbers...........

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06 Apr 2009, 00:57
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Re: prime numbers........... [#permalink]

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06 Apr 2009, 07:18
so p^2q^2 = 5(x)

Since both p and q are prime. q has to be 5 for this equation to be true.

so p^2 * 5 = 5(x)

So what is a multiple of 25 or 5^2(x) .

p^2 * 5 * 5 = 5*5(x)

= p^2q^2 ( since q = 5 )

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Re: prime numbers........... [#permalink]

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06 Apr 2009, 17:39
pbanavara wrote:
so p^2q^2 = 5(x)

where did you get that from ?
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Re: prime numbers........... [#permalink]

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06 Apr 2009, 22:16
pmenon wrote:
pbanavara wrote:
so p^2q^2 = 5(x)

where did you get that from ?

oops I meant p^2q = 5(x) From the question stem
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Re: prime numbers........... [#permalink]

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06 Apr 2009, 22:36
n = 5k. (n is a multiple of 5)
n = p^2.q (Where p and q are prime numbers)

This means, either p or q is 5 because n is a multiple of 5 afterall. NOTE: It can be EITHER of them.
Now we want multiple of 25 which is equal to 5^2. Thus:
Q = 5^2 . z (We need to find the multiple Q where z is a constant)

Now look at the options:
Can Q equal p^2? It could but we are not sure that p=5.
Can Q equal q^2? It could but we are not sure that q=5.
Can Q equal pq? No because we need 5 at least with an exponent of 2.
Can Q equal p^2q^2? YES. At least one of them is 5^2!!
Can Q equal p^3q? Again, what if q=5? In that case, we will be lacking one exponent of 5

Hence D it is
Hope this helps.
Re: prime numbers...........   [#permalink] 06 Apr 2009, 22:36
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# prime numbers...........

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