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# A set S contains numbers such that every element in the set

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Joined: 22 Jul 2014
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A set S contains numbers such that every element in the set  [#permalink]

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06 Aug 2014, 10:01
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Difficulty:

85% (hard)

Question Stats:

47% (01:53) correct 53% (02:15) wrong based on 152 sessions

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A set S contains numbers such that every element in the set S is greater than 1 and every element is a factor of both $$(3^7)*(5^3)*(7^2)*(11^7)$$ and $$(2^3)*(3^3)*(5^6)(7)$$. What is the maximum possible number of unique elements in S?

A) 43
B) 30
C) 29
D) 31
E) 32

Source: 4gmat
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Re: A set S contains numbers such that every element in the set  [#permalink]

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06 Aug 2014, 11:31
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If every number is a factor of both $$(3^7)*(5^3)*(7^2)*(11^7)$$ and $$(2^3)*(3^3)*(5^6)(7)$$, it should be a factor of their greatest common factor.
$$gcd((3^7)*(5^3)*(7^2)*(11^7), (2^3)*(3^3)*(5^6)(7))=3^3*5^3*7$$

To calculate the number of factors of certain number we need to find it's prime factorization (already done), add to each power 1 and multiply them.
Number of unique factors=(3+1)(3+1)(1+1)=4*4*2=32.

Since, every element is greater than we need subtract one factor 1: 32-1=31.

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Re: A set S contains numbers such that every element in the set  [#permalink]

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06 Aug 2014, 14:37
4
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alphonsa wrote:
A set S contains numbers such that every element in the set S is greater than 1 and every element is a factor of both $$(3^7)*(5^3)*(7^2)*(11^7)$$ and $$(2^3)*(3^3)*(5^6)(7)$$ .
What is the maximum possible number of unique elements in S?

A) 43
B) 30
C) 29
D) 31
E) 32

Source: 4gmat

Here is how it is with one simple formula and the rest a nominal logical approach.
Every element's got to be the factor of both the numbers, $$(3^7)*(5^3)*(7^2)*(11^7)$$ and $$(2^3)*(3^3)*(5^6)(7)$$

Do note the biggest number that can divide both these numbers should have those factors which can be found in both of these. (That's technically called GCD or the greatest common divisor) but even if I not talk about it technically, the greatest number I am seeking right now can only have a maximum of three 3's in its factorial expansions because while the first number has products of seven threes, the second number has only three 3's and hence I am restricted to use only three 3s. Similarly the greatest number can be this only: $$(3^3)*(5^3)*(7^1)$$

Here is the quickie formula I mentioned above... For a number which can be represented in the form of $$(a^b)*(c^d)*(e^f)$$ where all a, c and e are prime numbers and hence the prime factors of the number, the number of unique factors is given by $$(b+1)*(d+1)*(f+1)$$and so on.

Hence here the number of factors would be given by (3+1)*(3+1)*(1+1) = 4 * 4 * 2 = 32.
Since the numbers have to be all greater than 1, and the above result counts all factors including 1, we will subtract 1 from 32, hence the answer 31.

Let me know if this helps.

Regards,
Ketan

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Re: A set S contains numbers such that every element in the set  [#permalink]

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03 Jul 2016, 14:24
Basically calculate the gcd of given 2 numbers which is $$3^3$$ * $$5^3$$ * 7
According to this rule - ' To calculate the number of factors of certain number we need to find it's prime factorization (already done), add to each power 1 and multiply them.'
(3+1)(3+1)(1+1) = 32
but in this we have also considered 1 as a prime factor,whereas the question asks for factors greater than 1
so subtract 1 from 32
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Re: A set S contains numbers such that every element in the set  [#permalink]

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28 Jul 2017, 11:37
smyarga wrote:
If every number is a factor of both $$(3^7)*(5^3)*(7^2)*(11^7)$$ and $$(2^3)*(3^3)*(5^6)(7)$$, it should be a factor of their greatest common factor.
$$gcd((3^7)*(5^3)*(7^2)*(11^7), (2^3)*(3^3)*(5^6)(7))=3^3*5^3*7$$

To calculate the number of factors of certain number we need to find it's prime factorization (already done), add to each power 1 and multiply them.
Number of unique factors=(3+1)(3+1)(1+1)=4*4*2=32.

Since, every element is greater than we need subtract one factor 1: 32-1=31.

A very clear and good approach by you. Spasibo!
Re: A set S contains numbers such that every element in the set   [#permalink] 28 Jul 2017, 11:37
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