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M17-25

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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
[Reveal] Spoiler: OA

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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


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
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M17-25 [#permalink]

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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: M17-25 [#permalink]

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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: M17-25 [#permalink]

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New post 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!

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Re M17-25 [#permalink]

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New post 20 Aug 2016, 04:40
I think this is a high-quality question and I agree with explanation.
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Re: M17-25 [#permalink]

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New post 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,
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Re: M17-25 [#permalink]

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New post 08 Feb 2017, 14:38
confusing question
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Re: M17-25 [#permalink]

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New post 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: M17-25 [#permalink]

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New post 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: M17-25 [#permalink]

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New post 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 M17-25 [#permalink]

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New post 06 Apr 2018, 21:21
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.
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Re: M17-25 [#permalink]

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New post 06 Apr 2018, 23:51
Th3Gr80n3 wrote:
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.


The solution looks detailed enough. Please ask specific question or try alternative discussion here: https://gmatclub.com/forum/series-a-n-i ... 27175.html
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Re M17-25 [#permalink]

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New post 29 Apr 2018, 09:41
I think this is a high-quality 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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New post 29 Apr 2018, 09:43
patto wrote:
I think this is a high-quality 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


Could you please upload a screenshot? Thank you.
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Re: M17-25 [#permalink]

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That's the problem!!
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New post 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/m17-184131.html#p1453637


Yes 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 ?)

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Re: M17-25 [#permalink]

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New post 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/m17-184131.html#p1453637


Yes 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


Can you please try now? We fixed it. You might need to clean the cache first.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M17-25   [#permalink] 29 Apr 2018, 15:10

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