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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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New post 09 Dec 2014, 15:30
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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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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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Collection of Questions:
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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.

Thanks !

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Re: M17-25   [#permalink] 17 Sep 2017, 08:39
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