Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Affiliations: CFA Institute (CFA Candiate), Canadian Institute of Chartered Accountants (CA Candiate), Chartered Alternative Investments Analysts Association (CAIA Candidate)

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).
_________________

Please KUDOS generously if you feel I've been of service.

Last edited by Sovjet on 15 Aug 2011, 07:21, edited 1 time in total.

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).
_________________

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

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?

Affiliations: CFA Institute (CFA Candiate), Canadian Institute of Chartered Accountants (CA Candiate), Chartered Alternative Investments Analysts Association (CAIA Candidate)

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.
_________________

Please KUDOS generously if you feel I've been of service.

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

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.

Affiliations: CFA Institute (CFA Candiate), Canadian Institute of Chartered Accountants (CA Candiate), Chartered Alternative Investments Analysts Association (CAIA Candidate)

statement 1 is sufficient... in statement 2..its a geometric progression with the multiplying factor of 5. BUT, we cant deduce about other terms... correct answer is A can we have the OA plz bschool83
_________________

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.
_________________

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

Show Tags

27 Oct 2015, 20:12

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

06 Nov 2016, 10:04

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...