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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.



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Re: M1725 [#permalink]
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12 Jun 2012, 01:33
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: M1725 [#permalink]
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12 Jun 2012, 05:27
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 nonintegers > \(a_1=1=2a_2=3a_3=...\) > \(a_2=\frac{1}{2}\), \(a_3=\frac{1}{3}\), \(a_4=\frac{1}{4}\), 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: M1725 [#permalink]
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12 Jun 2012, 10:12
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 nonintegers > \(a_1=1=2a_2=3a_3=...\) > \(a_2=\frac{1}{2}\), \(a_3=\frac{1}{3}\), \(a_4=\frac{1}{4}\), 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: M1725 [#permalink]
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17 Sep 2012, 21:14
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 nonintegers > \(a_1=1=2a_2=3a_3=...\) > \(a_2=\frac{1}{2}\), \(a_3=\frac{1}{3}\), \(a_4=\frac{1}{4}\), 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: M1725 [#permalink]
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18 Sep 2012, 01:16
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 nonintegers > \(a_1=1=2a_2=3a_3=...\) > \(a_2=\frac{1}{2}\), \(a_3=\frac{1}{3}\), \(a_4=\frac{1}{4}\), 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: M1725 [#permalink]
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09 Feb 2013, 06:15
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 nonintegers > \(a_1=1=2a_2=3a_3=...\) > \(a_2=\frac{1}{2}\), \(a_3=\frac{1}{3}\), \(a_4=\frac{1}{4}\), 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: M1725 [#permalink]
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10 Feb 2013, 03:19
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 nonintegers > \(a_1=1=2a_2=3a_3=...\) > \(a_2=\frac{1}{2}\), \(a_3=\frac{1}{3}\), \(a_4=\frac{1}{4}\), 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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Re: M1725 [#permalink]
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07 Nov 2013, 21:41
Bunuel wrote: COULD be true, not MUS be true). 'MUS' typo is even in the answer to this question (M1725)










