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GRE Math Challenge #40-product of two or more of the prime

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GRE Math Challenge #40-product of two or more of the prime [#permalink] New post 19 Sep 2014, 03:25
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Question Stats:

50% (00:44) correct 50% (01:17) wrong based on 20 sessions
How many positive integers can be expressed as a product of two or more of the prime numbers 5,7,11 & 13 if no one product is to include the same prime factor more than one?

A. 8
B. 9
C. 10
D. 11
E. 12
[Reveal] Spoiler: OA

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Re: GRE Math Challenge #40-product of two or more of the prime [#permalink] New post 18 Sep 2017, 23:49
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4c1 +4c2 1 4c1
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Re: GRE Math Challenge #40-product of two or more of the prime [#permalink] New post 19 Sep 2017, 05:35
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Pushkar96 wrote:
4c1 +4c2 1 4c1


You probably meant \(4C2 + 4C3 + 4C4 = 6 + 4 + 1 = 11\). Thus, the answer is D
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Re: GRE Math Challenge #40-product of two or more of the prime [#permalink] New post 05 Sep 2018, 10:05
CAn anyone explain me how?
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Re: GRE Math Challenge #40-product of two or more of the prime [#permalink] New post 07 Sep 2018, 15:52
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IAMRV wrote:
CAn anyone explain me how?


Positive integers can be expressed as a product of two or more of the prime numbers 5,7,11 & 13 if no one product is to include the same prime factor more than once.

We can say that a number can have all the 4 prime factors i.e. \(5 \times 7 \times 11 \times 13 =5005\). This is 1 number.

Now a number can be formed by 3 numbers i.e. we need to find to 3 numbers from 4 order of section is not important = \(\frac{4!}{3! \times 1!}= 4\) numbers. Also written as \(C^{4}_{1}\).

Similarly a number can be formed using only 2 primes i.e. we need to find to " numbers from 4 order of section is not important = \(\frac{4!}{2! \times 2!}= 6\) numbers. Also written as \(C^{4}_{2}\).

Hence summing all the 3 cases. \(6+4+1=11\)
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Re: GRE Math Challenge #40-product of two or more of the prime   [#permalink] 07 Sep 2018, 15:52
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