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redundant prime

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redundant prime [#permalink] New post 13 Jun 2011, 07:45
Hi, I almost have a grasp on the concept, just need a little confirmation.

Q: If 24 is a factor of H and 28 is a factor of K, must 21 be a factor of HK?

A:
[Reveal] Spoiler:
Yes.....


The primes of H: 3,2,2,2...
The primes of K: 7,2,2...

Are the (2,2) in K redundant because they might be carried over from the (2,2) of H?

Please explain. Thx.
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Re: redundant prime [#permalink] New post 13 Jun 2011, 07:50
All the factors of both h and k must be
factors of the product, hk. Hence, the factors of hk include 2, 2, 2, 2,
2,3, and 7. Therefore, 21 is a factor of hk.
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Re: redundant prime [#permalink] New post 13 Jun 2011, 07:54
They might look redundant but in order to include 7 and 3, we have to take all the factors of h and k
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Re: redundant prime [#permalink] New post 13 Jun 2011, 07:55
subhashghosh wrote:
All the factors of both h and k must be
factors of the product, hk. Hence, the factors of hk include 2, 2, 2, 2,
2,3, and 7. Therefore, 21 is a factor of hk.


I know that. But my question is specifically regarding redundant primes.
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Re: redundant prime [#permalink] New post 13 Jun 2011, 08:54
386390 wrote:
subhashghosh wrote:
All the factors of both h and k must be
factors of the product, hk. Hence, the factors of hk include 2, 2, 2, 2,
2,3, and 7. Therefore, 21 is a factor of hk.


I know that. But my question is specifically regarding redundant primes.


Answer this:
Is HK divisible by 9?
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Re: redundant prime [#permalink] New post 13 Jun 2011, 09:12
i figured out what i was doing wrong:

if the question was: j/6=int & j/4=int

then there is a redundant prime involved but since in my question here the two variables are DIFFERENT (H,K), there is no redundant prime.
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Re: redundant prime   [#permalink] 13 Jun 2011, 09:12
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