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29 Apr 2018, 17:49
Now It's working fine! Ty Bunuel!!!



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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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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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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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04 Aug 2018, 00:17



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29 Aug 2018, 19:44
In statement 2 . I dont understand why we are only using a[1]=1 . Since a1 is positive integer than it is possible that a1=2. That will give us 2=2*a[2] hence a[2]=1 .Hence statement 2 is not always correct.



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29 Aug 2018, 20:15



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11 Sep 2018, 03:46
Bunuel wrote: 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 Hi Bunuel, Amazing questions  one doubt here, why would be say option 2 as Could Be True ?  isnt this as well a Must be True answer? TIA



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11 Sep 2018, 04:03
NidSha wrote: Bunuel wrote: 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 Hi Bunuel, Amazing questions  one doubt here, why would be say option 2 as Could Be True ?  isnt this as well a Must be True answer? TIA II COULD be true but it's not ALWAYS true. For example, if \(a_1=2\), then \(a_2=1\), so in this case \(a_1\) is NOT the only integer in the sequence.
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Re: M1725
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11 Oct 2018, 11:52
For statement 2
a1=2 and a2=1; so 1*a1=2*a2 in this case.
And both a1 and a2 are integers..
then wouldnt the statement a1 is the only integer be false?



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13 Oct 2018, 05:10
Bunuel wrote: 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. brunel , I am not able to find similar problems like this. Can you help ?







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