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Re M2008 [#permalink]
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16 Sep 2014, 01:08
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Re: M2008 [#permalink]
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How is statement 1 sufficient?
If we pick numbers for statement 1 we can get two different answers. if x=2 and y=3, then 1003=97, Answer NO if x=2 and y=9, then 1009=81, Answer YES
Am I missing something?



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28 Mar 2016, 22:16



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Re: M2008 [#permalink]
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23 May 2017, 06:21
Dear Sir,
I am having a doubt which might sound silly. Hence my apologies.
From the question, i understand is ((10^x)y)/9?
Is it okay if I rewrite this way: (10^x/9)y/9 since they will have comment denominator?
Is that case, since x and y are positive, wouldn't the stem of the question itself denotes that it is not divisible by 9?



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Re: M2008 [#permalink]
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23 May 2017, 06:34
Manoraaju wrote: Dear Sir,
I am having a doubt which might sound silly. Hence my apologies.
From the question, i understand is ((10^x)y)/9?
Is it okay if I rewrite this way: (10^x/9)y/9 since they will have comment denominator?
Is that case, since x and y are positive, wouldn't the stem of the question itself denotes that it is not divisible by 9? Yes, you can rewrite \(\frac{10^xy}{9}\) as \(\frac{10^x}{9}\frac{y}{9}\). But how can you say that \(\frac{10^x}{9}\frac{y}{9}\) will not be an integer? For example, if x = 1 and y = 1, then \(\frac{10^x}{9}\frac{y}{9}=1\)
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Re: M2008 [#permalink]
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04 Sep 2017, 08:30
Does this rule hold true for any integer? Quote: (something not divisible by 3)  (something divisible by 3) = (something not divisible by 3)



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04 Sep 2017, 08:49
Galiya wrote: Does this rule hold true for any integer? Quote: (something not divisible by 3)  (something divisible by 3) = (something not divisible by 3) Yes. WHEN THE SUM OR THE DIFFERENCE OF NUMBERS IS A MULTIPLE OF AN INTEGER1. If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):Example: \(a=6\) and \(b=9\), both divisible by 3 > \(a+b=15\) and \(ab=3\), again both divisible by 3. 2. If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 > \(a+b=11\) and \(ab=1\), neither is divisible by 3. 3. If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):Example: \(a=5\) and \(b=4\), neither is divisible by 3 > \(a+b=9\), is divisible by 3 and \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), both are divisible by 4. For more on this check: 5. Divisibility/Multiples/Factors 6. Remainders For other subjects: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative Megathread
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Re: M2008 [#permalink]
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04 Sep 2017, 08:59
Galiya wrote: Does this rule hold true for any integer? Quote: (something not divisible by 3)  (something divisible by 3) = (something not divisible by 3) Yeah this always holds true! Sent from my iPhone using GMAT Club Forum mobile app



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I am struggling a bit to understand option B. How are we approaching it?
Statement (2) by itself is insufficient. If \(y = 1\), the answer is "yes"; if \(y = 2\), the answer is "no".



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22 Sep 2017, 07:25
Shiprasingh1100 wrote: I am struggling a bit to understand option B. How are we approaching it?
Statement (2) by itself is insufficient. If \(y = 1\), the answer is "yes"; if \(y = 2\), the answer is "no". If y = 1, then 10^x + 1 could be 11, 101, 1001, ... so as given in the second statement, not divisible by 9. In this case, 10^x  1 will be 9, 99, 999, ... so divisible by 9. Answer YES. If y = 2, then 10^x + 1 could be 12, 102, 1002, ... so as given in the second statement, not divisible by 9. In this case, 10^x  1 will be 8, 98, 998, ... so NOT divisible by 9. Answer NO.
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Re: M2008 [#permalink]
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22 Sep 2017, 09:22
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Shiprasingh1100 It is stated in the question that x is positive integer i.e. x can be 1,2,3,4,5,6, and so on .So, 10^x will have have respective value 10,100,1000 and so on. Thus in 2. when y=1 9,99,999, and so on will always be divisible by 9 whereas when y=2 it will not always be divisible by 9. Hope you get it . Kudos if you like my reply



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Re: M2008 [#permalink]
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22 Sep 2017, 20:00
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Shiprasingh1100 wrote: I am struggling a bit to understand option B. How are we approaching it?
Statement (2) by itself is insufficient. If \(y = 1\), the answer is "yes"; if \(y = 2\), the answer is "no". The question is asking if 10^xy is divisible by 9. Since, for x=y=1; The answer is Yes and for x=2, y=9; The answer is No. Hence the option is not by itself sufficient.



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Re: M2008 [#permalink]
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25 Sep 2017, 10:39
oirfan wrote: How is statement 1 sufficient?
If we pick numbers for statement 1 we can get two different answers. if x=2 and y=3, then 1003=97, Answer NO if x=2 and y=9, then 1009=81, Answer YES
Am I missing something? Even I make such mistakes, be careful. How should one avoid such mistakes?










