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# Does the sequence a1, a2, a3, ..., an, ... contain an infini

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Does the sequence a1, a2, a3, ..., an, ... contain an infini  [#permalink]

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Updated on: 11 Jul 2013, 01:30
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Does the sequence a1, a2, a3, ..., an, ... contain an infinite number of terms that are divisible by 20?

(1) a1 = 5 and an = 4(5^(n – 1)) for all integers n ≥ 2.

(2) a2 = 20, a4 = 500, a5 = 2,500, and a6 = 12,500.

Originally posted by bschool83 on 14 Aug 2011, 09:36.
Last edited by Bunuel on 11 Jul 2013, 01:30, edited 1 time in total.
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Updated on: 15 Aug 2011, 08:21
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Statement (1):

a1 is obviously not divisible by 20. However, all integers n ≥ 2 are divisible by 20.

To ensure that a number is divisible by 20, a number should be divisible by 4 and 5 - a complete set of factors of 20. We could also choose to examine 1,2,10,20 as factors, but 4 and 5 suit our purposes here. We can see that the expression suggests that an is a product of both 5 and 4 as suggested by the 4 and 5 (factors) in the equation $$4(5^(^n^-^1^))$$. We can thus conclude that the number is divisible by 20.

Infinite = Infinite - Any Finite Number (in this case, a1)

This allows us to conclude that the set does indeed include an infinite number of terms divisible by 20. - Sufficient

Statment (2):

Here we note that each term in the sequence a2 = 20, a4 = 500, a5 = 2,500, and a6 = 12,500 is divisible by 20, as each term is simply a multiple of 20.

However, we can not infer based on this statement that this pattern continues, as we have no indication that this is a pattern. It's possible that only a2, a4, a5 and a6 follow this pattern, while the remaining (infinite number of) terms are random/not divisible by 20. Without more information, we cannot be certain. - Insufficient

Statement (1) is sufficient, while Statement (2) is not. Therefore, answer (A).
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Originally posted by Sovjet on 14 Aug 2011, 21:35.
Last edited by Sovjet on 15 Aug 2011, 08:21, edited 1 time in total.
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15 Aug 2011, 00:50
I will go with (A) here.

Statement (1): an = 4*5^(n-1) is divisible by 20 for all integers n>=2 is an identity. Therefore using this alone, we can say that the number of terms in the series divisible by 20 is infinite. Sufficient.

Statement (2): a2 = 20, a4 = 500, a5 = 2,500, and a6 = 12,500. Using this statement, we cannot infer the remaining terms of the series apart from the ones given. They may be divisible by 20, or may not be. If they are divisible by 20, and there is an infinite number of them, the series has an infinite number of terms divisible by 20. If they are not divisible by 20, then the number of terms divisible by 20, in this series, is finite. We do not know which case holds. Insufficient.

Therefore the answer should be (A).
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15 Aug 2011, 02:00
+1 for D,

We can infer from 2 that it is a GP where an = a(n-1) * 5 and since the starting term is 20, it is divisible by 20,
the next term is 100, then 500, then 2500 so on..

since all these terms going further are divisible by 100 they should also be divisible by 20..... so i also believe it is D
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15 Aug 2011, 02:31
Sovjet wrote:
Statement (1):

a1 is obviously not divisible by 20. However, all integers n ≥ 2 are divisible by 20.

To ensure that a number is divisible by 20, a number should be divisible by 4 and 5 - a complete set of factors of 20. We could also choose to examine 1,2,10,20 as factors, but 4 and 5 suit our purposes here. We can see that the expression suggests that an is a product of both 5 and 4 as suggested by the 4 and 5 (factors) in the equation $$4(5^(^n^-^1^))$$. We can thus conclude that the number is divisible by 20.

Infinite = Infinite - Any Finite Number (in this case, a1)

This allows us to conclude that the set does indeed include an infinite number of terms divisible by 20. - Sufficient

Statment (2):

Here we note that each even "an" is "$$a(n-2)*5$$". Thus, we can conclude that any number initially divisible by 20 will continue to be divisible regardless of the subsequent multiples. In other words, multiples of 20 (each time by a factor of 5) will always be divisible by 20.

It's possible that odd "an"s (for which we have no definition) are not divisible by 20, but that is relevant. Assume that all odd "an"s are not divisible, and you could still conclude that:

Infinite = Infinite - $$\frac{1}{2}$$ Infinite

This allows us to conclude that the set does indeed include an infinite number of terms divisible by 20. - Sufficient

Statement (1) and (2) are both sufficient. Therefore, answer (D).

In statement 1, a1=5 is definately not divisible by 20 though the remaining terms are..
shouldn't that make stmt 1 insufficient as not all the terms are divisible by 20?
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15 Aug 2011, 08:16
DeeptiM wrote:
In statement 1, a1=5 is definately not divisible by 20 though the remaining terms are..
shouldn't that make stmt 1 insufficient as not all the terms are divisible by 20?

The question isn't whether all numbers in the set are divisible by 20. The question asks whether there is an infinite number of terms in the set divisible by 20.(Infinite - 1 = Infinite)

I see where GyanOne is coming from though. I jumped to the conclusion that the set continues in a pattern after a6 - whereas this may not be the case. The answer might indeed be (A).

Upon reflection, I think GyanOne is correct. Edited initial post.
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15 Aug 2011, 08:57
IMO D...
A- no dispute here-SUFFICIENT
B- there is sequence in which the terms a2,a4,a5 seem to be related..
a2-20
a3-20*5=100
a4=20*5*5=500
a5=20*5*5*5=
so,each term has the number 20 in it.//so SUFFICIENT
hence ,IMO D
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15 Aug 2011, 20:03
Sovjet wrote:
DeeptiM wrote:
In statement 1, a1=5 is definately not divisible by 20 though the remaining terms are..
shouldn't that make stmt 1 insufficient as not all the terms are divisible by 20?

The question isn't whether all numbers in the set are divisible by 20. The question asks whether there is an infinite number of terms in the set divisible by 20.(Infinite - 1 = Infinite)

I see where GyanOne is coming from though. I jumped to the conclusion that the set continues in a pattern after a6 - whereas this may not be the case. The answer might indeed be (A).

Upon reflection, I think GyanOne is correct. Edited initial post.

Ouch!! missed the "infinite no of terms" part..

Thnx Sovjet!!
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16 Aug 2011, 16:36
I am Convinced with Option A as answer.
Statement 2 is not convincing as it does not say the series will continue forever.

Any convincing solutions to prove solution is D please.?
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17 Aug 2011, 10:36
Very tricky, we have to be very careful...
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17 Aug 2011, 10:40
I think it's probably time for bschool to post the OA.
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07 Oct 2011, 01:03
OA would be useful please, but as formulated, the question is a pretty straightforward A
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07 Oct 2011, 06:44
A..
B is not sufficient because of the "infinity" requirement
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Updated on: 08 Oct 2011, 08:58
1
statement 1 is sufficient...
in statement 2..its a geometric progression with the multiplying factor of 5.
BUT,
we cant deduce about other terms...
can we have the OA plz bschool83
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Originally posted by Marcab on 08 Oct 2011, 08:51.
Last edited by Marcab on 08 Oct 2011, 08:58, edited 1 time in total.
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08 Oct 2011, 08:57
siddharthasingh wrote:
statement 1 is sufficient...
in statement 2..its a geometric progression with the multiplying factor of 5.
hence,
a2=20, a3=100, a4=500, a5=2500, a6=12500, a7=62500..and so on..
clearly each and every element is divisible by 20..
IMO D..
can we have the OA plz bschool83
..
i guess i missed out something..
we cant deduce that other terms must be following the GP..
hence i would eliminate my previous answer..
the correct answer should be A.
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09 Oct 2011, 12:56
IMO D.

Statement 2 looks to be a geometric series and all the terms are a multiple of 4 and 5. for n>=2.
Why is statement 2 not sufficient?

Please can you post the OA?

Thanks
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Re: Does the sequence a1, a2, a3, ..., an, ... contain an infini  [#permalink]

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22 Oct 2017, 08:27
I would go with A just because the first term (a1) is 5, and so this statement is sufficient, however the answer to the question of the statement is "no" since 5 is not divisible by 20
Re: Does the sequence a1, a2, a3, ..., an, ... contain an infini &nbs [#permalink] 22 Oct 2017, 08:27
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