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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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16 Sep 2014, 01:01
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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 [#permalink]
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09 Dec 2014, 16:07



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Re: M1725 [#permalink]
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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 [#permalink]
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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 [#permalink]
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08 Feb 2017, 14:38
confusing question



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Re: M1725 [#permalink]
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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 [#permalink]
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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.
Thanks !



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17 Sep 2017, 08:39



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Re: M1725 [#permalink]
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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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06 Apr 2018, 21:21
I think this is a poorquality question and the explanation isn't clear enough, please elaborate.



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29 Apr 2018, 09:41
I think this is a highquality question. HI!! I''ve got a problema when I want to see the explanations. I always find that numbers are all in the same zone one in front of the other. It's a mess. I cannot distinguish any number! Please provide a solution to this problem! I can send a picture if you want



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29 Apr 2018, 10:06
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That's the problem!!
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29 Apr 2018, 10:14



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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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