a data from gmat club

This topic has expert replies
Legendary Member
Posts: 1119
Joined: Fri May 07, 2010 8:50 am
Thanked: 29 times
Followed by:3 members

a data from gmat club

by diebeatsthegmat » Sun Aug 21, 2011 12:25 am
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.

i think the answer should be D instead of A :( dun understand why it is A. can you please explain?

User avatar
Master | Next Rank: 500 Posts
Posts: 102
Joined: Tue Mar 08, 2011 3:42 am
Thanked: 2 times

by arashyazdiha » Sun Aug 21, 2011 3:25 am
Well for considering the answer D , you should prove that each statement alone holds.
maybe reading the first statement gives you a bias. but second statement shows few ones and does not cover the whole sequence. so How can you sat that it contains an infinite number of items ....?

Legendary Member
Posts: 2789
Joined: Tue Jul 26, 2011 12:19 am
Location: Chennai, India
Thanked: 206 times
Followed by:43 members
GMAT Score:640

by GmatKiss » Mon Aug 22, 2011 2:30 am
I agree with diebeatsthegmat, it seems to be a D for me!!

Experts help needed :)

thanks,
GK

Master | Next Rank: 500 Posts
Posts: 381
Joined: Wed May 19, 2010 10:15 pm
Thanked: 41 times
Followed by:2 members

by AbhiJ » Mon Aug 22, 2011 3:54 am
D should be the answer.

Using GP rule

T(n) = a * r ^(n-1)
We know T(4), T(5), T(6), which gives the series.

Note: However we are not given with the above formula and the series can be some other series as well. If we were given the above formula then we could prove the above series using Principal of Mathematical Induction. However we are not given the formula hence GP for all n, cannot be proved.

Having said that this is out of the scope of GMAT Prep, and if this question appears on the GMAT the most likely answer would be D.

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 273
Joined: Tue Sep 21, 2010 5:37 am
Location: Raleigh, NC
Thanked: 154 times
Followed by:74 members
GMAT Score:770

by Whitney Garner » Mon Aug 22, 2011 8:31 am
AbhiJ wrote:D should be the answer.

Using GP rule

T(n) = a * r ^(n-1)
We know T(4), T(5), T(6), which gives the series.

Note: However we are not given with the above formula and the series can be some other series as well. If we were given the above formula then we could prove the above series using Principal of Mathematical Induction. However we are not given the formula hence GP for all n, cannot be proved.

Having said that this is out of the scope of GMAT Prep, and if this question appears on the GMAT the most likely answer would be D.
Be careful, we only know that this is a GP from statement 1.

Statement (2) gives us NO information about the "formula" for this sequence - so it might not even have one. Therefore, we cannot know the values for any other terms except for those given, so this is Insufficient. Those values given ARE divisible by 20, but we cannot know about any of the others.

Statement (1) is sufficient because it tells us that EVERY term in the sequence is divisible by 4 and then 5 to some power, EXCEPT, the first term in the sequence is NOT divisible by 20 - therefore it is sufficient = the answer to the question "Does the sequence {a1, a2, a3, ..., an, ...} contain an infinite number of terms that are divisible by 20?" would actually be NO.

Extended FYI - this question seems fairly un-GMAT-like in that Statement (1) gives a1=5, a number clearly not divisible by 20, and still provides the geometric calculation for the rest of the series.

:)
Whit
Whitney Garner
GMAT Instructor & Instructor Developer
Manhattan Prep

Contributor to Beat The GMAT!

Math is a lot like love - a simple idea that can easily get complicated :)

Senior | Next Rank: 100 Posts
Posts: 47
Joined: Wed Sep 03, 2008 11:18 pm
Thanked: 5 times
Followed by:1 members

by top_business_2011 » Mon Aug 22, 2011 8:41 am
AbhiJ wrote:D should be the answer.

Using GP rule

T(n) = a * r ^(n-1)
We know T(4), T(5), T(6), which gives the series.

Note: However we are not given with the above formula and the series can be some other series as well. If we were given the above formula then we could prove the above series using Principal of Mathematical Induction. However we are not given the formula hence GP for all n, cannot be proved.

Having said that this is out of the scope of GMAT Prep, and if this question appears on the GMAT the most likely answer would be D.
The answer is definitely A.
Let's see:

Required: Does the sequence contain an infinite number of terms divisible by 20?

Statement 1: Since al = 5, which is not divisible by 20, we can respond to the question saying,"No!" Hence, sufficient. [I think there is something wrong with statement 1, because even without 'an = 4(5^(n - 1)) for all integers n ≥ 2' the answer can be determined.]

Statement 2: We have no information about whether the sequence contains more than 6 terms.
the sequence may be just: { 20,100,500,2500, 12,500} In this case, the answer to our basic question is,"No!". However, if the sequence is continuous, then the answer is "Yes!". Hence, Insufficient.

Except a minor flow, in my opinion, in formulating statement 1, the problem is nice and is by no means outside the scope of the real GMAT. I have seen some questions,from trusted sources,testing similar concept.
Here is one of them: If P is a set of integers and 3 is in P, is every positive multipe of 3 in P?

1) For any integer in P, the sum of 3 and that integer is aslo in P.

2) For any integer in P, that integer minus 3 is also in P. [Ans. A]

Master | Next Rank: 500 Posts
Posts: 381
Joined: Wed May 19, 2010 10:15 pm
Thanked: 41 times
Followed by:2 members

by AbhiJ » Tue Aug 23, 2011 12:55 am
top_business_2011 wrote:
The answer is definitely A.
Let's see:

Required: Does the sequence contain an infinite number of terms divisible by 20?

Statement 1: Since al = 5, which is not divisible by 20, we can respond to the question saying,"No!" Hence, sufficient. [I think there is something wrong with statement 1, because even without 'an = 4(5^(n - 1)) for all integers n ≥ 2' the answer can be determined.]

Statement 2: We have no information about whether the sequence contains more than 6 terms.
the sequence may be just: { 20,100,500,2500, 12,500} In this case, the answer to our basic question is,"No!". However, if the sequence is continuous, then the answer is "Yes!". Hence, Insufficient.

Except a minor flow, in my opinion, in formulating statement 1, the problem is nice and is by no means outside the scope of the real GMAT. I have seen some questions,from trusted sources,testing similar concept.
Here is one of them: If P is a set of integers and 3 is in P, is every positive multipe of 3 in P?

1) For any integer in P, the sum of 3 and that integer is aslo in P.

2) For any integer in P, that integer minus 3 is also in P. [Ans. A]
I see your point that the series can be only 5 terms. But the question stem talks about a continous series (a1, a2, a3.....an)? Moreover in option B we are not said that it doesnot contain infinite terms. So one is free to assume that its a continous series by common logic. Its the type of ambiguity you will not find in any OG problem.

Master | Next Rank: 500 Posts
Posts: 381
Joined: Wed May 19, 2010 10:15 pm
Thanked: 41 times
Followed by:2 members

by AbhiJ » Tue Aug 23, 2011 1:05 am
Whitney Garner wrote: Be careful, we only know that this is a GP from statement 1.

Statement (2) gives us NO information about the "formula" for this sequence - so it might not even have one. Therefore, we cannot know the values for any other terms except for those given, so this is Insufficient. Those values given ARE divisible by 20, but we cannot know about any of the others.

Extended FYI - this question seems fairly un-GMAT-like in that Statement (1) gives a1=5, a number clearly not divisible by 20, and still provides the geometric calculation for the rest of the series.

:)
Whit
Have one doubt when you say that statement 2 doesnot talks about any series formula.

If its posed as a question in problem solving section as a series question
20, 100, 500, 2500, 12500, .....

Asking for the next term should the answer not be based on Gometric Progression.

Senior | Next Rank: 100 Posts
Posts: 47
Joined: Wed Sep 03, 2008 11:18 pm
Thanked: 5 times
Followed by:1 members

by top_business_2011 » Tue Aug 23, 2011 8:01 am
AbhiJ wrote:
top_business_2011 wrote:
The answer is definitely A.
Let's see:

Required: Does the sequence contain an infinite number of terms divisible by 20?

Statement 1: Since al = 5, which is not divisible by 20, we can respond to the question saying,"No!" Hence, sufficient. [I think there is something wrong with statement 1, because even without 'an = 4(5^(n - 1)) for all integers n ≥ 2' the answer can be determined.]

Statement 2: We have no information about whether the sequence contains more than 6 terms.
the sequence may be just: { 20,100,500,2500, 12,500} In this case, the answer to our basic question is,"No!". However, if the sequence is continuous, then the answer is "Yes!". Hence, Insufficient.

Except a minor flow, in my opinion, in formulating statement 1, the problem is nice and is by no means outside the scope of the real GMAT. I have seen some questions,from trusted sources,testing similar concept.
Here is one of them: If P is a set of integers and 3 is in P, is every positive multipe of 3 in P?

1) For any integer in P, the sum of 3 and that integer is aslo in P.

2) For any integer in P, that integer minus 3 is also in P. [Ans. A]
I see your point that the series can be only 5 terms[If that's so, then it is NOT infinite]. But the question stem talks about a continous series (a1, a2, a3.....an...)?[ Well, here I agree with you that the question is somewhat ambiguous] Moreover in option B we are not said that it doesnot contain infinite terms. So one is free to assume that its a continous series by common logic.[Never restrict your domain, unless you're expressly told to do so! Just as we shouldn't assume a number to be an integer only because it is easier to deal with] Its the type of ambiguity you will not find in any OG problem.[ You can say that again!]