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Bunuel
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M17-25 01 Jul 2023, 03:00
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SrinivasPrabhu
Hi Bunuel,

Why cannot option 1 be true for this scenario? What about the case where the set is {1, 1/2, 1/3, .....1/98, 1/99, 1/100}
Isn't i * ai always 1? Which makes 2 * a100 = 2 and a99 + a98 = 2 as well?
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Your scenario will occur when \(a_1=1\). In this case, the sequence will indeed be {1, 1/2, 1/3, ..., 1/98, 1/99, 1/100}. Note that for any term in the sequence, the product of that term and its index (its position number in the sequence) is always 1: \(1*a_1=2*a_2=3*a_3=...=100*a_{100}=1\). However, \(2*a_{100}=a_{99}+a_{98}\) does not hold as \(2*a_{100}=\frac{1}{50}\), whereas \(a_{99}+a_{98}=\frac{1}{98}+\frac{1}{99}=\frac{197}{9702}\neq 2\).

Furthermore, observe that \(i*a_1\) is not always 1. For example, when \(a_1=2\), the sequence becomes {2, 2/2, 2/3, 2/4, 2/5, 2/6, ...} and in this scenario, \(i*a_1=2\), not 1.

Hope it helps.
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M17-25 23 Jul 2023, 17:48
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Hey

So in 'could be true' questions, the statement may or may not be true?

Jai
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M17-25 24 Jul 2023, 00:54
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JAIDEEP27
Hey

So in 'could be true' questions, the statement may or may not be true?

Jai
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"MUST BE TRUE" ("ALWAYS TRUE"/"IS TRUE") questions:
    These questions ask which statement is always true for every valid set of numbers. If you can find just one valid set of numbers where a statement is not true, it means the statement is not always true and therefore not the correct answer.

    So, for 'MUST BE TRUE' questions, the plug-in method is good for eliminating an option, but it does not provide a 100% guarantee that an option is always true.

    For "MUST BE TRUE" questions, when using the plug-in method, if you find that more than one option appears to be correct for a particular number or set of numbers, try using different numbers to double-check. Reevaluate only those options that were previously considered correct.

"COULD BE TRUE" questions:
    The questions that ask which of the following statements could be true are different. If you can demonstrate that a statement is true for a specific set of numbers, it implies that the statement could be true and therefore is a correct answer.

You can practice more questions of this type HERE.

Hope it helps.
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I think this is a high-quality question and I agree with explanation.
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I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I don't understand why the first option cannot be true if a100 = a99 = a98; then why isn't a100 = a98 and 2a100 = a99 + a98
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M17-25 23 Oct 2024, 02:25
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Bunuel
Official 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 \(a_1\), \(a_2\), \(a_3\), ... has the following properties: \(i*a_i=j*a_j\) and \(a_1=\text{positive integer}\). Thus, \(1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=\text{positive integer}\).

We need to determine which of the given options can occur (note that the question asks which of the following COULD be true, not MUST be true).

I. \(2a_{100}=a_{99}+a_{98}\). Since \(100a_{100}=99a_{99}=98a_{98}\), then \(2a_{100}=\frac{100}{99}a_{100}+\frac{100}{98}a_{100}\). Divide 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. Given that \(a_1=\text{positive integer}=n*a_n\), then \(a_n=\frac{\text{positive integer}}{n}=\text{positive number}\). Therefore, this option is always true.


Answer: D
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I don't understand why the first option cannot be true if a100 = a99 = a98; then why isn't a100 = a98 and 2a100 = a99 + a98
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We are not given that \(a_{100} = a_{99} = a_{98} =...\), instead we are given \(1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=\text{positive integer}\), so \(100a_{100}=99a_{99}=98a_{98}\).
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I don't understand how can this question be a 705-805 level. In my opinion it's a 605-655 (max) type of question
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M17-25 11 Mar 2025, 09:44
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Ale300503
I don't understand how can this question be a 705-805 level. In my opinion it's a 605-655 (max) type of question
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We do not assign the difficulty level manually. The difficulty level of a question on the site is determined automatically based on various parameters collected from users' attempts, such as the percentage of correct answers and the time taken to answer the question. You can find the difficulty level of a question and its related statistics in the first post. So, the difficulty level of this question is 705-805 (Hard) Level Level based on the timer attempts from the users.
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why (i,j) are (1,2) why i is not equal to j... it does not say they can't be same please explain..... and if you any tips on how can i improve qaunt score please share some tips even after studying quant for 3 months i am still struggling.

Bunuel
Official 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 \(a_1\), \(a_2\), \(a_3\), ... has the following properties: \(i*a_i=j*a_j\) and \(a_1=\text{positive integer}\). Thus, \(1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=\text{positive integer}\).

We need to determine which of the given options can occur (note that the question asks which of the following COULD be true, not MUST be true).

I. \(2a_{100}=a_{99}+a_{98}\). Since \(100a_{100}=99a_{99}=98a_{98}\), then \(2a_{100}=\frac{100}{99}a_{100}+\frac{100}{98}a_{100}\). Divide 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. Given that \(a_1=\text{positive integer}=n*a_n\), then \(a_n=\frac{\text{positive integer}}{n}=\text{positive number}\). Therefore, this option is always true.


Answer: D
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M17-25 11 Mar 2025, 10:50
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MAN11
why (i,j) are (1,2) why i is not equal to j... it does not say they can't be same please explain..... and if you any tips on how can i improve qaunt score please share some tips even after studying quant for 3 months i am still struggling.

Bunuel
Official 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 \(a_1\), \(a_2\), \(a_3\), ... has the following properties: \(i*a_i=j*a_j\) and \(a_1=\text{positive integer}\). Thus, \(1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=\text{positive integer}\).

We need to determine which of the given options can occur (note that the question asks which of the following COULD be true, not MUST be true).

I. \(2a_{100}=a_{99}+a_{98}\). Since \(100a_{100}=99a_{99}=98a_{98}\), then \(2a_{100}=\frac{100}{99}a_{100}+\frac{100}{98}a_{100}\). Divide 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. Given that \(a_1=\text{positive integer}=n*a_n\), then \(a_n=\frac{\text{positive integer}}{n}=\text{positive number}\). Therefore, this option is always true.


Answer: D
Show more

We are not given that \(i = j\); instead, we are given that \(i * a_i = j * a_j\) for any pair of positive integers \((i, j)\). This equation defines a relationship between different terms in the sequence.

This is a hard question, and it's okay if you don't get it on the first try. I suggest reviewing the discussion again carefully.
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M17-25 23 Mar 2025, 07:15
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Bunuel
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why (i,j) are (1,2) why i is not equal to j... it does not say they can't be same please explain..... and if you any tips on how can i improve qaunt score please share some tips even after studying quant for 3 months i am still struggling.

Bunuel
Official 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 \(a_1\), \(a_2\), \(a_3\), ... has the following properties: \(i*a_i=j*a_j\) and \(a_1=\text{positive integer}\). Thus, \(1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=\text{positive integer}\).

We need to determine which of the given options can occur (note that the question asks which of the following COULD be true, not MUST be true).

I. \(2a_{100}=a_{99}+a_{98}\). Since \(100a_{100}=99a_{99}=98a_{98}\), then \(2a_{100}=\frac{100}{99}a_{100}+\frac{100}{98}a_{100}\). Divide 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. Given that \(a_1=\text{positive integer}=n*a_n\), then \(a_n=\frac{\text{positive integer}}{n}=\text{positive number}\). Therefore, this option is always true.


Answer: D

We are not given that \(i = j\); instead, we are given that \(i * a_i = j * a_j\) for any pair of positive integers \((i, j)\). This equation defines a relationship between different terms in the sequence.

This is a hard question, and it's okay if you don't get it on the first try. I suggest reviewing the discussion again carefully.
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Bunuel I think it's better to mention explicitly that i is not equal to j since the question says "for any pair of positive integers (i,j)". So according to this, any pair should satisfy the sequence including i=j right?
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M17-25 23 Mar 2025, 07:23
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siddhantvarma

Bunuel I think it's better to mention explicitly that i is not equal to j since the question says "for any pair of positive integers (i,j)". So according to this, any pair should satisfy the sequence including i=j right?
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No. For i = j we get \(i * a_i = i * a_i\), which is also obviously true.
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I don’t quite agree with the solution. the seq. only gives info about the proportion of the as, a1 doesnt have to be = to 1, it has to be =2 times a2, but if a1 is equal to 2 for example, then a1=2.a. 2=2.a2. then a2=1 and it is a integer.
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M17-25 09 Sep 2025, 23:52
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nobiseos
I don’t quite agree with the solution. the seq. only gives info about the proportion of the as, a1 doesnt have to be = to 1, it has to be =2 times a2, but if a1 is equal to 2 for example, then a1=2.a. 2=2.a2. then a2=1 and it is a integer.
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You are missing a point and not reading the question carefully enough. Also, it seems you skipped the previous discussion of this question on the earlier pages. The key is that we are not asked which of the statements must be true, we are asked which of the following could be true. That means if we can find even one configuration, one set of values, that makes a statement true, then it qualifies as part of the correct answer.

The solution shows that statements II and III could be true for some specific values. This doesn’t mean they are always true, but they are true for some sets of values. I suggest you review the full discussion and study it thoroughly.
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I like the solution - it’s helpful. It is a high-quality question, and the explanation is correct
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I like the solution - it’s helpful.
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I like the solution - it’s helpful.
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I did not quite understand the solution. Can not understand the 1st point
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I did not quite understand the solution. Can not understand the 1st point
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Have you checked the two pages of the discussion? The question is elaborated there in quite a bit of detail. Please review.
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Why can't we have the following sequence: 1 , 1/2 , 1/3 , 1/4 ...
In this case 2a100 will be equal to a99 + a98
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