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Re M3006 [#permalink]
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16 Sep 2014, 01:45
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, 15:56
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, 04:42



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10 Feb 2015, 06:14
Thanks!



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Re: M3006 [#permalink]
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01 Jun 2016, 19:57
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, 22:51
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, 06:47
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, 07:50
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, 12:41
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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11 Oct 2016, 06:14



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Re: M3006 [#permalink]
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23 Sep 2017, 04:37
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, 04:42
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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Re: M3006 [#permalink]
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06 Feb 2018, 04:23
Hi Bunuel,
I'm super confused as to how statement 2 works...the examples given in statement 2: (4,8) and (2,5) seem to be subsets of set A, yet (4,8) is said to be NO, and (2,5) is said to be YES. Can you please help me on this point? thanks.



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Re: M3006 [#permalink]
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06 Feb 2018, 05:28
ttaiwo wrote: Hi Bunuel,
I'm super confused as to how statement 2 works...the examples given in statement 2: (4,8) and (2,5) seem to be subsets of set A, yet (4,8) is said to be NO, and (2,5) is said to be YES. Can you please help me on this point? thanks. Set A = {2, 5, 11, 17, 23, 29, 41, ...} (distinct prime numbers which are 2 more than a multiple of 3). The question asks whether set B is a subset of set A. If from (2) set B is {4, 8}, then the answer is NO: {4, 8} is NOT a subset of {2, 5, 11, 17, 23, 29, 41, ...} If from (2) set B is {2, 5}, then the answer is YES: {2, 5} IS a subset of { 2, 5, 11, 17, 23, 29, 41, ...} Hope it's clear.
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Re: M3006 [#permalink]
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06 Feb 2018, 08:44
Bunuel wrote: ttaiwo wrote: Hi Bunuel,
I'm super confused as to how statement 2 works...the examples given in statement 2: (4,8) and (2,5) seem to be subsets of set A, yet (4,8) is said to be NO, and (2,5) is said to be YES. Can you please help me on this point? thanks. Set A = {2, 5, 11, 17, 23, 29, 41, ...} (distinct prime numbers which are 2 more than a multiple of 3). The question asks whether set B is a subset of set A. If from (2) set B is {4, 8}, then the answer is NO: {4, 8} is NOT a subset of {2, 5, 11, 17, 23, 29, 41, ...} If from (2) set B is {2, 5}, then the answer is YES: {2, 5} IS a subset of { 2, 5, 11, 17, 23, 29, 41, ...} Hope it's clear. Thanks  It's now clear.










