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M17-25

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Re: M17-25  [#permalink]

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New post 03 Aug 2018, 23:17
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salilgupta4180 wrote:
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?


Notice that the question is which of the following COULD be true, not MUS be true. II COULD be true if a1 = 1. In this case a1 will be the only integer in the sequence.
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Re: M17-25  [#permalink]

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New post 29 Aug 2018, 18:44
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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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Re: M17-25  [#permalink]

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New post 29 Aug 2018, 19:15
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shuvodip04 wrote:
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.


Notice that the question is which of the following COULD be true, not MUS be true. II COULD be true if a1 = 1. In this case a1 will be the only integer in the sequence.
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Re: M17-25  [#permalink]

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New post 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 non-integers, 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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Re: M17-25  [#permalink]

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New post 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 non-integers, 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: M17-25  [#permalink]

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New post 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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New post 11 Oct 2018, 19:07
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Re: M17-25  [#permalink]

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New post 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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Re: M17-25  [#permalink]

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New post 20 Nov 2018, 06:54
If a1 is positive, how do we know if j or n is poistive?
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Re: M17-25 &nbs [#permalink] 20 Nov 2018, 06:54

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