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For any positive integer n, the length of n

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Manager
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For any positive integer n, the length of n [#permalink]

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New post 19 Dec 2008, 17:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75 = 3 x 5 x 5. How many two-digit positive integers have a length of 6?

(A) None
(B) One
(C) Two
(D) Three
(E) Four


Can someone please give me an approach to how to solve this kind of problem?

thank you.
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Re: Prime factors [#permalink]

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New post 19 Dec 2008, 17:40
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Before I go ahead, it looks like you have selected A:None and the answer is C:TWO.

I am not too sure whether you have understood the concept of prime factors. I will explain it anyway, just to refresh my memory. :)

'Prime factors are prime numbers that divide an integer, without leaving a remainder'
eg: 6 = 2 * 3, so 6 has TWO prime factors.

Now coming to the problem here, it is asking us to find integers starting from 11 to 99 that can be divided by 6 prime factors(all of them could be same or different)

Lets pick the smallest of all : 2
1. 2^6 = 2*2*2*2*2*2 = 64 - ONE

Lets use 2 and 3 now
2. 2^5 *3 = 2*2*2*2*2*3 = 96 - TWO

Now lets try to use 4 2's and 2 3's
2*2*2*2*3*3 = 144

That means, we can't go any further.

So, the answer has to be C: TWO

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Kudos [?]: 47 [0], given: 0

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Re: Prime factors [#permalink]

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New post 19 Dec 2008, 18:06
Thanks, LiveStronger.

thank you for the excellent explanation. I couldn't think of a solid approach to the problem when I got to it on my practice test. Didn't even think about starting with the lowest prime!!:)

thanks again.

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Re: Prime factors   [#permalink] 19 Dec 2008, 18:06
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For any positive integer n, the length of n

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