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Re M3006 [#permalink]
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16 Sep 2014, 00:45
Official Solution:According to the stem set A = {2, 5, 11, 17, 23, 29, 41, ...} (1) Set B consist of two positive integers whose product is 10. 10 can be broken into the product of two integers in two ways: 10 = 1*10 and 10 = 2*5. If set B = {1, 10}, then the answer to the question is NO but if set B = {2, 5}, then the answer to the question is YES. Not sufficient. (2) The product of reciprocals of all elements in set B is a terminating decimal. If set B = {4, 8}, then the answer to the question is NO but if set B = {2, 5}, then the answer to the question is YES. Not sufficient. (1)+(2) Both possible sets from (1) meet the condition stated in the second statement: \(\frac{1}{1}*\frac{1}{10}=\frac{1}{10}=0.1= terminating \ decimal\) and \(\frac{1}{2}*\frac{1}{5}=\frac{1}{10}=0.1= terminating \ decimal\). Thus we still have two sets, which give two different answers to the question. Not sufficient. Answer: E
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Re: M3006 [#permalink]
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31 Jan 2015, 14:56
Bunuel,
How come you included 2 in the set A? Should the set A start with 5 instead of 2?



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Re: M3006 [#permalink]
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01 Feb 2015, 03:42



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Re: M3006 [#permalink]
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10 Feb 2015, 05:14
Thanks!



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Re: M3006 [#permalink]
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01 Jun 2016, 18:57
Is there any easy way to tell if a fraction is a terminating decimal?



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Re: M3006 [#permalink]
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01 Jun 2016, 21:51
glochou wrote: Is there any easy way to tell if a fraction is a terminating decimal? Theory:Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^2\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\). Note that if denominator already has only 2s and/or 5s then it doesn't matter whether the fraction is reduced or not. For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal. We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced. Check Terminating and Recurring Decimals Problems in our Special Questions Directory.
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Re M3006 [#permalink]
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06 Aug 2016, 05:47
I think this is a highquality question and I agree with explanation.



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Re: M3006 [#permalink]
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21 Sep 2016, 06:50
These kind of questions teach you to very carefully approach every single GMAT problem. Zero is a multiple of every single integer  it is an often overlooked property of zero and it is the key to solve this question.



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Re: M3006 [#permalink]
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10 Oct 2016, 11:41
I want to ask a question
if a subset includes more than one copy of a member of the original set. Then is it still considered a subset? for example A= 1,2,3 and if B =1.2,2. Would B be a subset of A?



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Re: M3006 [#permalink]
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11 Oct 2016, 05:14



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Re: M3006 [#permalink]
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23 Sep 2017, 03:37
Hi Bunuel,
Set A consists of all distinct prime numbers which are 2 more than a multiple of 3. Can you please explain this bit clearly.
I can see the set as ( 2,5,11, etc) , here 3 and 7 are missing, but 7 is not a multiple of 3. Explain this point too.



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Re: M3006 [#permalink]
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23 Sep 2017, 03:42
msk0657 wrote: Hi Bunuel,
Set A consists of all distinct prime numbers which are 2 more than a multiple of 3. Can you please explain this bit clearly.
I can see the set as ( 2,5,11, etc) , here 3 and 7 are missing, but 7 is not a multiple of 3. Explain this point too. Set A consists of all distinct prime numbers which are 2 more than a multiple of 3, so primes which are 2 more than a multiple of 3, so {a multiple of 3} + 2 = prime: {2, 5, 11, 17, 23, 29, 41, ...} How does 3 or 7 satisfy {a multiple of 3} + 2 = prime?
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