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M17-25

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M17-25 [#permalink] New post 11 Jun 2012, 22:51
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

55% (03:47) correct 45% (01:39) wrong based on 11 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

Please, explain the the solution.
[Reveal] Spoiler: OA
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Re: M17-25 [#permalink] New post 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] New post 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

Please, explain the the solution.


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.

Answer: D.

Hope it's clear.
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Re: M17-25 [#permalink] New post 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.

Answer: D.

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] New post 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

Please, explain the the solution.




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.

Answer: D.

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] New post 18 Sep 2012, 00:16
1
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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

Please, explain the the solution.




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.

Answer: D.

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] New post 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

Please, explain the the solution.


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.

Answer: D.

Hope it's clear.



can you please explain me option A. i am totally confused with it
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Re: M17-25 [#permalink] New post 10 Feb 2013, 02:19
1
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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

Please, explain the the solution.


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.

Answer: D.

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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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] New post 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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