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# M17-25

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Math Expert
Joined: 02 Sep 2009
Posts: 51215
Re: M17-25  [#permalink]

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03 Aug 2018, 23:17
1
salilgupta4180 wrote:
For the statment 2, instead of taking a1 as 1 if we take it as 2 then there woudl be 2 integers in this case. Hence the statement can be false aswell
a2 = a1/2 => 2/2 => 1

So why option 2 is correct?

Notice that the question is which of the following COULD be true, not MUS be true. II COULD be true if a1 = 1. In this case a1 will be the only integer in the sequence.
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Re: M17-25  [#permalink]

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29 Aug 2018, 18:44
1
In statement 2 . I dont understand why we are only using a[1]=1 . Since a1 is positive integer than it is possible that a1=2. That will give us 2=2*a[2] hence a[2]=1 .Hence statement 2 is not always correct.
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29 Aug 2018, 19:15
1
shuvodip04 wrote:
In statement 2 . I dont understand why we are only using a[1]=1 . Since a1 is positive integer than it is possible that a1=2. That will give us 2=2*a[2] hence a[2]=1 .Hence statement 2 is not always correct.

Notice that the question is which of the following COULD be true, not MUS be true. II COULD be true if a1 = 1. In this case a1 will be the only integer in the sequence.
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Re: M17-25  [#permalink]

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11 Sep 2018, 02:46
Bunuel wrote:
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 $$a_1$$, $$a_2$$, $$a_3$$, ... 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

Hi Bunuel, Amazing questions - one doubt here, why would be say option 2 as Could Be True ? - isnt this as well a Must be True answer?

TIA
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Re: M17-25  [#permalink]

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11 Sep 2018, 03:03
NidSha wrote:
Bunuel wrote:
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 $$a_1$$, $$a_2$$, $$a_3$$, ... 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

Hi Bunuel, Amazing questions - one doubt here, why would be say option 2 as Could Be True ? - isnt this as well a Must be True answer?

TIA

II COULD be true but it's not ALWAYS true. For example, if $$a_1=2$$, then $$a_2=1$$, so in this case $$a_1$$ is NOT the only integer in the sequence.
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Re: M17-25  [#permalink]

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11 Oct 2018, 10:52
For statement 2-

a1=2 and a2=1;
so 1*a1=2*a2 in this case.

And both a1 and a2 are integers..

then wouldnt the statement a1 is the only integer be false?
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Re: M17-25  [#permalink]

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11 Oct 2018, 19:07
vishaldd01 wrote:
For statement 2-

a1=2 and a2=1;
so 1*a1=2*a2 in this case.

And both a1 and a2 are integers..

then wouldnt the statement a1 is the only integer be false?

__________________________
Your doubt is addressed above.
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Re: M17-25  [#permalink]

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13 Oct 2018, 04:10
Bunuel wrote:
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.

brunel ,
I am not able to find similar problems like this. Can you help ?
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Re: M17-25  [#permalink]

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20 Nov 2018, 06:54
If a1 is positive, how do we know if j or n is poistive?
Re: M17-25 &nbs [#permalink] 20 Nov 2018, 06:54

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# M17-25

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