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# M17-25

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M17-25 [#permalink]  11 Jun 2012, 22:51
00:00

Difficulty:

5% (low)

Question Stats:

50% (02:50) correct 50% (01:39) wrong based on 10 sessions
Series 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 is possible?

I. 2A(100)=A(99)+A(98)
II. A(1) is the only integer in the series
III. The series does not contain negative numbers

I only
II only
I and III only
II and III only
I, II, and III

[Reveal] Spoiler: OA
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Re: M17-25 [#permalink]  12 Jun 2012, 00:33
i*A(i) = j*A(j)

Take i=1,
A(j) = A(1)/j
so,
A(2) = A(1)/2
A(3) = A(1)/3
.
.
.
The series is A(1), A(1)/2, A(1)/3...

I 2A(1)/100 = A(1)/99 + A(1)/98
2/100 = 1/99 + 1/98, this is definitely not true.

II IF A(1)=1, then all other numbers in the series will not be integers,
so II is possible.

III. Since A(1) is positive integer,
remaining numbers in the series will be positive only. This is always true.

So, II and III CAN be true.
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Re: M17-25 [#permalink]  12 Jun 2012, 04:27
Expert's post
RuslanMRF wrote:
Series 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 is possible?

I. 2A(100)=A(99)+A(98)
II. A(1) is the only integer in the series
III. The series does not contain negative numbers

I only
II only
I and III only
II and III only
I, II, and III

New edition of this question with a 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=positive \ integer, so 1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=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 be true.

II. a_1 is the only integer in the sequence. If a_1=1, then all other terms will be non-integers --> a_1=1=2a_2=3a_3=... --> 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=positive \ integer=n*a_n, then a_n=\frac{positive \ integer}{n}=positive \ number, hence this option is always true.

Hope it's clear.
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Re: M17-25 [#permalink]  12 Jun 2012, 09:12
Bunuel wrote:
New edition of this question with a 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=positive \ integer, so 1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=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 be true.

II. a_1 is the only integer in the sequence. If a_1=1, then all other terms will be non-integers --> a_1=1=2a_2=3a_3=... --> 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=positive \ integer=n*a_n, then a_n=\frac{positive \ integer}{n}=positive \ number, hence this option is always true.

Hope it's clear.

The new edition of question is clearer than the previous one.

I believe there is a typo in explanation of I.
It should be Not True.
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Re: M17-25 [#permalink]  17 Sep 2012, 20:14
Bunuel wrote:
RuslanMRF wrote:
Series 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 is possible?

I. 2A(100)=A(99)+A(98)
II. A(1) is the only integer in the series
III. The series does not contain negative numbers

I only
II only
I and III only
II and III only
I, II, and III

New edition of this question with a 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=positive \ integer, so 1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=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 be true.

II. a_1 is the only integer in the sequence. If a_1=1, then all other terms will be non-integers --> a_1=1=2a_2=3a_3=... --> 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=positive \ integer=n*a_n, then a_n=\frac{positive \ integer}{n}=positive \ number, hence this option is always true.

Hope it's clear.

Are we assuming that, since j comes after i in the alphabet, that j is equivalent to i+1? If so, it is much clearer to write n*a_n=(n+1)*a_(n+1) because i and j could have any relationship
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Re: M17-25 [#permalink]  18 Sep 2012, 00:16
1
KUDOS
Expert's post
dandarth1 wrote:
Bunuel wrote:
RuslanMRF wrote:
Series 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 is possible?

I. 2A(100)=A(99)+A(98)
II. A(1) is the only integer in the series
III. The series does not contain negative numbers

I only
II only
I and III only
II and III only
I, II, and III

New edition of this question with a 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=positive \ integer, so 1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=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 be true.

II. a_1 is the only integer in the sequence. If a_1=1, then all other terms will be non-integers --> a_1=1=2a_2=3a_3=... --> 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=positive \ integer=n*a_n, then a_n=\frac{positive \ integer}{n}=positive \ number, hence this option is always true.

Hope it's clear.

Are we assuming that, since j comes after i in the alphabet, that j is equivalent to i+1? If so, it is much clearer to write n*a_n=(n+1)*a_(n+1) because i and j could have any relationship

We are told that "i*a_i=j*a_j for any pair of positive integers (i, j)", so it has nothing to do whether j comes after i and we are not assuming that j=i+1.
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Re: M17-25 [#permalink]  09 Feb 2013, 05:15
Bunuel wrote:
RuslanMRF wrote:
Series 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 is possible?

I. 2A(100)=A(99)+A(98)
II. A(1) is the only integer in the series
III. The series does not contain negative numbers

I only
II only
I and III only
II and III only
I, II, and III

New edition of this question with a 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=positive \ integer, so 1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=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 be true.

II. a_1 is the only integer in the sequence. If a_1=1, then all other terms will be non-integers --> a_1=1=2a_2=3a_3=... --> 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=positive \ integer=n*a_n, then a_n=\frac{positive \ integer}{n}=positive \ number, hence this option is always true.

Hope it's clear.

can you please explain me option A. i am totally confused with it
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Posts: 15073
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Re: M17-25 [#permalink]  10 Feb 2013, 02:19
1
KUDOS
Expert's post
FTG wrote:
Bunuel wrote:
RuslanMRF wrote:
Series 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 is possible?

I. 2A(100)=A(99)+A(98)
II. A(1) is the only integer in the series
III. The series does not contain negative numbers

I only
II only
I and III only
II and III only
I, II, and III

New edition of this question with a 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=positive \ integer, so 1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=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 be true.

II. a_1 is the only integer in the sequence. If a_1=1, then all other terms will be non-integers --> a_1=1=2a_2=3a_3=... --> 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=positive \ integer=n*a_n, then a_n=\frac{positive \ integer}{n}=positive \ number, hence this option is always true.

Hope it's clear.

can you please explain me option A. i am totally confused with it

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]  07 Nov 2013, 20:41
Bunuel wrote:
COULD be true, not MUS be true).

'MUS' typo is even in the answer to this question (M17-25)
Re: M17-25   [#permalink] 07 Nov 2013, 20:41
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# M17-25

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