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Question Stats:
48% (02:18) correct 52% (02:11) wrong based on 309 sessions
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The sequence \(a_1\), \(a_2\), \(a_3\), ..., \(a_n\), ... is such that \(i*a_i=j*a_j\) for any pair of positive integers \((i, j)\). If \(a_1\) is a positive integer, which of the following could be true? I. \(2*a_{100}=a_{99}+a_{98}\) II. \(a_1\) is the only integer in the sequence III. The sequence does not contain negative numbers A. I only B. II only C. I and III only D. II and III only E. I, II, and III
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Re M1725
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16 Sep 2014, 01:01
Official Solution: The sequence \(a_1\), \(a_2\), \(a_3\), ..., \(a_n\), ... is such that \(i*a_i=j*a_j\) for any pair of positive integers \((i, j)\). If \(a_1\) is a positive integer, which of the following could be true? I. \(2*a_{100}=a_{99}+a_{98}\) II. \(a_1\) is the only integer in the sequence III. The sequence does not contain negative numbers A. I only B. II only C. I and III only D. II and III only E. I, II, and III Given that the sequence of numbers \(a_1\), \(a_2\), \(a_3\), ... have the following properties: \(i*a_i=j*a_j\) and \(a_1=\text{positive integer}\), so \(1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=\text{positive integer}\). We should determine whether the options given below can occur (notice that the question is which of the following COULD be true, not MUS be true). I. \(2a_{100}=a_{99}+a_{98}\). Since \(100a_{100}=99a_{99}=98a_{98}\), then \(2a_{100}=\frac{100}{99}a_{100}+\frac{100}{98}a_{100}\). Reduce by \(a_{100}\): \(2=\frac{100}{99}+\frac{100}{98}\) which is not true. Hence this option could NOT be true. II. \(a_1\) is the only integer in the sequence. If \(a_1=1\), then all other terms will be nonintegers, because in this case we would have \(a_1=1=2a_2=3a_3=...\), which leads to \(a_2=\frac{1}{2}\), \(a_3=\frac{1}{3}\), \(a_4=\frac{1}{4}\), and so on. Hence this option could be true. III. The sequence does not contain negative numbers. Since given that \(a_1=\text{positive integer}=n*a_n\), then \(a_n=\frac{\text{positive integer}}{n}=\text{positive number}\), hence this option is always true. Answer: D
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Hi Bunuel, I got this question on a GMAT Club test and even after reviewing it I'm having trouble understanding what is going on. The following deduction is giving me trouble: since 100a(sub)100= 99a(sub)99 =98a(sub)98, then 2a(sub)100 = [100][99/]a(sub)100 + [100][98/]a(sub)100. << Please see attachment below if this is not clear. I hope this makes sense. Cheers.
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Re: M1725
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09 Dec 2014, 16:07
Darknightw wrote: Hi Bunuel, I got this question on a GMAT Club test and even after reviewing it I'm having trouble understanding what is going on. The following deduction is giving me trouble: since 100a(sub)100= 99a(sub)99 =98a(sub)98, then 2a(sub)100 = [100][99/]a(sub)100 + [100][98/]a(sub)100. << Please see attachment below if this is not clear. I hope this makes sense. Cheers. From \(100a_{100}=99a_{99}\) > \(a_{99}=\frac{100}{99}a_{100}\); From \(100a_{100}=98a_{98}\) > \(a_{98}=\frac{100}{98}a_{100}\); So, option I. \(2a_{100}=a_{99}+a_{98}\) becomes: \(2a_{100}=\frac{100}{99}a_{100}+\frac{100}{98}a_{100}\). Hope it's clear.
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Re: M1725
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09 Dec 2014, 16:13
Bunuel,
Thanks for the quick response. Yes, it's now very clear. Can't believe I couldn't see that!
Thanks again!
Darknightw
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20 Aug 2016, 04:40
I think this is a highquality question and I agree with explanation.



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Re: M1725
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01 Sep 2016, 20:30
Hi Bunuel, Quick question. I solved the question as below. I assume that a(x) as a function, which reciprocates the integer x. so 100a100 = 100(1/100) = 1. Thus even negative integer say 1a(1) will give a value of 1 and will equal any other xa(x) value. Now I see that Stmt 1: 2 (1/100) = (1/99)+(1/98) Not Possible Stmt 2: a1 = 1/1 = 1; every thing else is 1/2; 1/1 ( still an integer) meaning no negative numbers in the sequence Stmt 3: As above the sequence may or may not contain negative numbers. Since the question stem says " Could be true" I chose Stmt 2 and 3 to be the right answer. Is my reasoning valid? Thanks, Arun



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Re: M1725
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08 Feb 2017, 14:38
confusing question



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Re: M1725
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15 May 2017, 07:01
The answer must be option D. There are some awesome explanations above. I did it using a slightly different method though. ai/aj=j/i express a100 , a99, & a98 in terms of a1. Now go to each and every statement. Statement 1  substitute in LHS,RHS. Solve and you get LHS not equal to RHS. Hence 1 is wrong Since 1 is wrong we are left with choices B and D. II is common to both. Jump on to statement III. III a1 is positive i & j are positive so any number of the form an will be positive. Hence III is true. The answer is option D
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Re: M1725
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17 Sep 2017, 07:11
Can anyone please provide an alternative solution of this problem. I have trouble understanding the official answer explanation.
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Re: M1725
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17 Sep 2017, 08:39
jyotipes21@gmail.com wrote: Can anyone please provide an alternative solution of this problem. I have trouble understanding the official answer explanation.
Thanks ! Check here: https://gmatclub.com/forum/seriesani ... 27175.html
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Re: M1725
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13 Feb 2018, 06:46
The word "could be" makes the difference. Otherwise, option II is not always correct. For instance, if a1 = 2 , there could be two integers in the sequence.



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Re: M1725
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29 Apr 2018, 10:06
That's the problem!!
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Re: M1725
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29 Apr 2018, 10:14
patto wrote: That's the problem!! Thank you. We are looking into this. Could you see posts in this discussion OK? For example, could you see this post: https://gmatclub.com/forum/m17184131.html#p1453637
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Re: M1725
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29 Apr 2018, 10:24
Bunuel wrote: patto wrote: That's the problem!! Thank you. We are looking into this. Could you see posts in this discussion OK? For example, could you see this post: https://gmatclub.com/forum/m17184131.html#p1453637Yes i can see all the posts but in all of them, the numbers are one in front the other in the same place. I found this problem in all the discussions that correspond to the questions of the same test(my first test ?) Posted from my mobile device
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Re: M1725
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29 Apr 2018, 15:10
patto wrote: Bunuel wrote: patto wrote: That's the problem!! Thank you. We are looking into this. Could you see posts in this discussion OK? For example, could you see this post: https://gmatclub.com/forum/m17184131.html#p1453637Yes i can see all the posts but in all of them, the numbers are one in front the other in the same place. I found this problem in all the discussions that correspond to the questions of the same test(my first test ?) Posted from my mobile deviceCan you please try now? We fixed it. You might need to clean the cache first.
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Re: M1725
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29 Apr 2018, 17:49
Now It's working fine! Ty Bunuel!!!
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Re: M1725
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19 May 2018, 12:36
Bunuel wrote: Official Solution:
II. \(a_1\) is the only integer in the sequence. If \(a_1=1\), then all other terms will be nonintegers, because in this case we would have \(a_1=1=2a_2=3a_3=...\), which leads to \(a_2=\frac{1}{2}\), \(a_3=\frac{1}{3}\), \(a_4=\frac{1}{4}\), and so on. Hence this option could be true. III. The sequence does not contain negative numbers. Since given that \(a_1=\text{positive integer}=n*a_n\), then \(a_n=\frac{\text{positive integer}}{n}=\text{positive number}\), hence this option is always true.
Why does the explanation for II say "could be true" and not "always true" (like it does for III)?



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Re: M1725
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19 May 2018, 14:01
aserghe1 wrote: Bunuel wrote: Official Solution:
II. \(a_1\) is the only integer in the sequence. If \(a_1=1\), then all other terms will be nonintegers, because in this case we would have \(a_1=1=2a_2=3a_3=...\), which leads to \(a_2=\frac{1}{2}\), \(a_3=\frac{1}{3}\), \(a_4=\frac{1}{4}\), and so on. Hence this option could be true. III. The sequence does not contain negative numbers. Since given that \(a_1=\text{positive integer}=n*a_n\), then \(a_n=\frac{\text{positive integer}}{n}=\text{positive number}\), hence this option is always true.
Why does the explanation for II say "could be true" and not "always true" (like it does for III)? The question asks: which of the following COULD be true. If an option is true, is possible, even for one sequence then it fits. I is not true for this sequence at all. So it's out. II COULD be true in certain case, so it fits. III is ALWAYS true, so it also fits. Hope it's clear.
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Re: M1725
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03 Aug 2018, 14:52
For the statment 2, instead of taking a1 as 1 if we take it as 2 then there woudl be 2 integers in this case. Hence the statement can be false aswell a2 = a1/2 => 2/2 => 1
So why option 2 is correct?







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