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:

In a certain sequence, every term after the first is determi [#permalink]

Show Tags

12 Jan 2012, 22:42

4

This post received KUDOS

14

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

25% (02:57) correct
75% (02:09) wrong based on 761 sessions

HideShow timer Statistics

In a certain sequence, every term after the first is determined by multiplying the previous term by an integer constant greater than 1. If the fifth term of the sequence is less than 1000, what is the maximum number of non-negative integer values possible for the first term?

Re: Non negative integers in a sequence [#permalink]

Show Tags

13 Jan 2012, 04:52

8

This post received KUDOS

1

This post was BOOKMARKED

enigma123 wrote:

In a certain sequence, every term after the first is determined by multiplying the previous term by an integer constant greater than 1. If the fifth term of the sequence is less than 1000, what is the maximum number of non-negative integer values possible for the first term? A) 60 B) 61 C) 62 D) 63 E) 64

Any idea on the concept and how to solve this please?

You can solve the problem using equations or take a logical route.

I'll do it by logical elimination. We are given the following: T1, T2, T3..T5 are 5 terms of a sequence, where T2 = K*T1 , T3=K*T2 [K> 1 & T5 < 1000]

To get the MAXIMUM number of non-negative integer values for T1 , we have to take the value of K as low as possible and K> 1 (given) therefore let us take K=2. Looking at the options we can figure out that value of T1 will be b/w 60-64.

Lets us start by the maximum value of 64. T1=64, T2=128......T5=1024 (but T5 < 1000), hence E is out T1=63,T2=126........T5=1008 ,Hence D is out T1=62, T2=64.........T5= 992. This works .Thus all values from 1 to 62 will work . But we are told that T1 is a non-negative integer, so T1=0 (zero) will also work. Thus 0 to 62 values will work . Therefore total number of values equals 63 .Option D.

Alternatively equations can be formed as shown above T (n+1) = K* Tn and T5 <1000. substutuite for T5=1000 and get T1 (T1=62.5) ,but T1 has to be integer therefore T1=62 ,then the method as above. Hope this helps

What is the source of the problem and the answer?
_________________

In a certain sequence, every term after the first is determined by multiplying the previous term by an integer constant greater than 1. If the fifth term of the sequence is less than 1000, what is the maximum number of non-negative integer values possible for the first term? A) 60 B) 61 C) 62 D) 63 E) 64

Any idea on the concept and how to solve this please?

Given sequence: \(x\); \(x*r\); \(x*r^2\); \(x*r^3\); \(x*r^4<1,000\) (where x is the first term and r is the constant greater than 1).

To maximize the # of non-negative integer values possible for \(x\), we should minimize the value of \(r\) and since \(r=integer>1\) then \(r=2\). (General rule for such kind of problems: to maximize one quantity, minimize the others and to minimize one quantity, maximize the others.)

Thus, \(x*2^4<1,000\) -->\(x<\frac{1,000}{16}=62,5\) --> as the first term must be a non-negative integer then: \(x_{max}=62\) and \(x_{min}=0\) --> total of 63 values possible for the first term x: {0, 1, 2, ..., 62}.

Re: Non negative integers in a sequence [#permalink]

Show Tags

14 Jan 2012, 03:30

[quote="Bunuel] Thus, \(x*2^4<1,000\) -->\(x<\frac{1,000}{16}=62,5\) --> as the first term must be a non-negative integer then: \(x_{max}=62\) and \(x_{min}=0\) --> total of 63 values possible for the first term x: {0, 1, 2, ..., 62}.

Answer: D.[/quote]

Bunnel here you have taken x_min as 0, and here we are given to find the max no of non negative integers. Since 0 is neither +ve nor -ve, so do we still have to take 0??? I took x_min as 1.

Thus, \(x*2^4<1,000\) -->\(x<\frac{1,000}{16}=62,5\) --> as the first term must be a non-negative integer then: \(x_{max}=62\) and \(x_{min}=0\) --> total of 63 values possible for the first term x: {0, 1, 2, ..., 62}.

Answer: D.

Bunnel here you have taken x_min as 0, and here we are given to find the max no of non negative integers. Since 0 is neither +ve nor -ve, so do we still have to take 0??? I took x_min as 1.

We are told that the first term of the sequence is a non-negative integer, so yes, the first term could equal to zero.

In this case we'll have the sequence with all numbers equal to zero: \(x_{min}=0\); \(x*r=0\); \(x*r^2=0\); \(x*r^3=0\); \(x*r^4=0<1,000\), ... (By the way for this scenario \(r\) could be any integer)

For the case when the first term is 62 (and \(r=2\)) the sequence will be: \(x_{max}=62\); \(x*r=124\); \(x*r^2=248\); \(x*r^3=496\); \(x*r^4=992<1,000\).

As you can see the first term can take all integer values from 0 to 62, inclusive: {0, 1, 2, ..., 62}, so total of 63 values.

Re: In a certain sequence, every term after the first is determi [#permalink]

Show Tags

08 Feb 2012, 02:33

Hi Bunuel , I have a doubt regarding the use of '0' as a first term . If we see the wording of the qs, it says - 'every term after the first is determined by multiplying the previous term by an integer constant greater than 1.' If all the term of the sequence are '0' - then the statement is wrong - which is not possible . Why so ? Say all the term are indeed '0' - then to get next term we can multiply anything with the previous term not specifically 'an integer constant greater than 1' as told by the statement .

In this line of explanation the right answer should be C> 62 and not D>63 .

Hi Bunuel , I have a doubt regarding the use of '0' as a first term . If we see the wording of the qs, it says - 'every term after the first is determined by multiplying the previous term by an integer constant greater than 1.' If all the term of the sequence are '0' - then the statement is wrong - which is not possible . Why so ? Say all the term are indeed '0' - then to get next term we can multiply anything with the previous term not specifically 'an integer constant greater than 1' as told by the statement .

In this line of explanation the right answer should be C> 62 and not D>63 .

Please clarify . Thanks, VCG.

I see your point. But, it's kind of other way around.

If the first term is 0 then the first five terms will be {0, 0, 0, 0, 0} and this set is perfectly OK. Yes, in this case, r can be any integer, not necessarily greater than 1, though if is is greater than 1, then the set still holds true.

Re: In a certain sequence, every term after the first is determi [#permalink]

Show Tags

05 Jun 2013, 06:52

In a certain sequence, every term after the first is determined by multiplying the previous term by an integer constant greater than 1. If the fifth term of the sequence is less than 1000, what is the maximum number of non-negative integer values possible for the first term? A) 60 B) 61 C) 62 D) 63 E) 64

In order to max the first number we need to min the constant integer. The min integer value greater than 1 is 2, so lets take constant as 2. Another thing we know is that the fith term is less than 1000. Our sequence is the following X, 2X, 4X, 8X, 16X. Basically 16X<1000 ---> x<62.5, the closest integer value is 62. If the first integer value is 62 the fith will be less than 1000, that means that all the nonnegative integers less than 62 will fit into our conditions. Overall there are 0...62=63 integer values. So the answer is D.
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Re: In a certain sequence, every term after the first is determi [#permalink]

Show Tags

03 Aug 2013, 22:32

What if the question stated that we need to find the minimum number of values that a can take?

So, again => ar^4<1000

we need to maximise r^4: if we pick r =4, then we have r^4=256. then a can take 4 values. (a<3.96) so a can be 3,2,1,0 However, if we take r=5, then a ^4=625 and a can take 2 values i. 0 as a<1000/625 => a<1.xy(xy is some decimal value) so can either be one or a can be 0

Re: In a certain sequence, every term after the first is determi [#permalink]

Show Tags

03 Aug 2013, 22:47

1

This post received KUDOS

12bhang wrote:

What if the question stated that we need to find the minimum number of values that a can take?

So, again => ar^4<1000

we need to maximise r^4: if we pick r =4, then we have r^4=256. then a can take 4 values. (a<3.96) so a can be 3,2,1,0 However, if we take r=5, then a ^4=625 and a can take 2 values i. 0 as a<1000/625 => a<1.xy(xy is some decimal value) so can either be one or a can be 0

so minimum 2 values.

we cannot take r =6 because 6^4>1000

Am i right in my reasoning?

IF the question asks that, the minimum no of NON-NEGATIVE values that a can take is 1, for a=0. Then, no matter how large the value of r is,it will really not make any difference.

Re: In a certain sequence, every term after the first is determi [#permalink]

Show Tags

03 Jul 2014, 16:52

dentobizz wrote:

enigma123 wrote:

In a certain sequence, every term after the first is determined by multiplying the previous term by an integer constant greater than 1. If the fifth term of the sequence is less than 1000, what is the maximum number of non-negative integer values possible for the first term? A) 60 B) 61 C) 62 D) 63 E) 64

Any idea on the concept and how to solve this please?

You can solve the problem using equations or take a logical route.

I'll do it by logical elimination. We are given the following: T1, T2, T3..T5 are 5 terms of a sequence, where T2 = K*T1 , T3=K*T2 [K> 1 & T5 < 1000]

To get the MAXIMUM number of non-negative integer values for T1 , we have to take the value of K as low as possible and K> 1 (given) therefore let us take K=2. Looking at the options we can figure out that value of T1 will be b/w 60-64.

Lets us start by the maximum value of 64. T1=64, T2=128......T5=1024 (but T5 < 1000), hence E is out T1=63,T2=126........T5=1008 ,Hence D is out T1=62, T2=64.........T5= 992. This works .Thus all values from 1 to 62 will work . But we are told that T1 is a non-negative integer, so T1=0 (zero) will also work. Thus 0 to 62 values will work . Therefore total number of values equals 63 .Option D.

Alternatively equations can be formed as shown above T (n+1) = K* Tn and T5 <1000. substutuite for T5=1000 and get T1 (T1=62.5) ,but T1 has to be integer therefore T1=62 ,then the method as above. Hope this helps

What is the source of the problem and the answer?

WITH YOUR LOGICAL ELIMINATION HOW DID YOU GET T5 = 1024? and T2 = 128?

r can be 2,3,4,5 only . but all the solutions mentioned in the post only assume r = 2 .

however if we take the above mentioned values for r ... we will get even more values of a.

eg : if r =2 => a can have 63 values as explained above. if r =3 => a can have another 12 values etc isn't it??

please let me know if im wrong.

thanks.

It seems that you misinterpreted the question. The question asks: what is the maximum number of non-negative integer values possible for the first term?

We have that: \(x*r^4<1,000\) (where x is the first term and r is the constant greater than 1).

To maximize the # of non-negative integer values possible for \(x\), we should minimize the value of \(r\) and since \(r=integer>1\) then \(r=2\).

Re: In a certain sequence, every term after the first is determi [#permalink]

Show Tags

06 Aug 2014, 22:48

Bunuel wrote:

enigma123 wrote:

In a certain sequence, every term after the first is determined by multiplying the previous term by an integer constant greater than 1. If the fifth term of the sequence is less than 1000, what is the maximum number of non-negative integer values possible for the first term? A) 60 B) 61 C) 62 D) 63 E) 64

Any idea on the concept and how to solve this please?

Given sequence: \(x\); \(x*r\); \(x*r^2\); \(x*r^3\); \(x*r^4<1,000\) (where x is the first term and r is the constant greater than 1).

To maximize the # of non-negative integer values possible for \(x\), we should minimize the value of \(r\) and since \(r=integer>1\) then \(r=2\). (General rule for such kind of problems: to maximize one quantity, minimize the others and to minimize one quantity, maximize the others.)

Thus, \(x*2^4<1,000\) -->\(x<\frac{1,000}{16}=62,5\) --> as the first term must be a non-negative integer then: \(x_{max}=62\) and \(x_{min}=0\) --> total of 63 values possible for the first term x: {0, 1, 2, ..., 62}.

Answer: D.

Thanks, I came up with 62, should have read the question more carefully
_________________

......................................................................... +1 Kudos please, if you like my post

Re: In a certain sequence, every term after the first is determi [#permalink]

Show Tags

07 Aug 2014, 10:16

Since we must find the maximum number of values, we must minimize the constant of multiplication i.e-2.

If we plug in 64, we end up with 1024 as the 5th term. If we plug in 63, we end up with 1008 as the 5th term. If we plug in 62, we end up with 992 as the 5th term.

So we can have 62 values from 1 and we must remember that 0 is a possible value too.

Therefore in total, there can be 63 such possible values.

Re: In a certain sequence, every term after the first is determi [#permalink]

Show Tags

18 Jun 2015, 08:22

enigma123 wrote:

In a certain sequence, every term after the first is determined by multiplying the previous term by an integer constant greater than 1. If the fifth term of the sequence is less than 1000, what is the maximum number of non-negative integer values possible for the first term?

A) 60 B) 61 C) 62 D) 63 E) 64

I guess it is 63. when k=2, the fifth term is 16x62<1000 and a can assume zero.

gmatclubot

Re: In a certain sequence, every term after the first is determi
[#permalink]
18 Jun 2015, 08:22

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Term 1 has begun. If you're confused, wondering what my post on the last 2 official weeks was, that was pre-term. What that means is that the school...