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03 Aug 2018, 23:17



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29 Aug 2018, 18:44
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, 19:15



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11 Sep 2018, 02:46
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, 03:03
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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11 Oct 2018, 10:52
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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11 Oct 2018, 19:07



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13 Oct 2018, 04:10
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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20 Nov 2018, 06:54
If a1 is positive, how do we know if j or n is poistive?







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