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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]  19 Sep 2014, 03:25
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Question Stats:

54% (00:50) correct 45% (01:17) wrong based on 22 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]  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]  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]  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]  07 Sep 2018, 15:52
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Expert's post
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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