Find all School-related info fast with the new School-Specific MBA Forum

It is currently 25 Oct 2014, 19:58

Close

GMAT Club Daily Prep

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

In a certain sequence, every term after the first is determi

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
1 KUDOS received
Director
Director
avatar
Status: Preparing for the 4th time -:(
Joined: 25 Jun 2011
Posts: 560
Location: United Kingdom
Concentration: International Business, Strategy
GMAT Date: 06-22-2012
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 13

Kudos [?]: 534 [1] , given: 217

In a certain sequence, every term after the first is determi [#permalink] New post 12 Jan 2012, 21:42
1
This post received
KUDOS
4
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

21% (02:27) correct 79% (02:08) wrong based on 417 sessions
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
[Reveal] Spoiler: OA

_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610 :-(

Manager
Manager
avatar
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 178
Location: India
Concentration: Finance, International Business
WE: Information Technology (Investment Banking)
Followers: 3

Kudos [?]: 22 [0], given: 1

Re: Non negative integers in a sequence [#permalink] New post 13 Jan 2012, 02:15
Ans is C

From the statement we can make an eq as a_n = a_(n-1)*k where n>1 and k>1

given the fifth term is less than 1000 i.e. a_5 <1000

to solve this I first take a_5 = 1000
to get the first term as max I take k=2 and using the above equation get a_1 = 62.5

since a_5<1000 so a_1 = 62
4 KUDOS received
Verbal Forum Moderator
Verbal Forum Moderator
User avatar
Joined: 23 Jul 2010
Posts: 490
Followers: 43

Kudos [?]: 445 [4] , given: 249

GMAT ToolKit User Premium Member CAT Tests
Re: Non negative integers in a sequence [#permalink] New post 13 Jan 2012, 03:52
4
This post received
KUDOS
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?
_________________

How to CHOOSE your Business School
Thanks = Kudos. Kudos are appreciated

Rules for posting on the verbal forum
When you post a question Pls. Provide its source & TAG your questions
Avoid posting from unreliable sources such as 1000 series.


Last edited by dentobizz on 13 Jan 2012, 04:18, edited 1 time in total.
Expert Post
6 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23422
Followers: 3618

Kudos [?]: 28981 [6] , given: 2874

Re: Non negative integers in a sequence [#permalink] New post 13 Jan 2012, 04:17
6
This post received
KUDOS
Expert's post
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?



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.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Director
Director
avatar
Status: Preparing for the 4th time -:(
Joined: 25 Jun 2011
Posts: 560
Location: United Kingdom
Concentration: International Business, Strategy
GMAT Date: 06-22-2012
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 13

Kudos [?]: 534 [0], given: 217

Re: Non negative integers in a sequence [#permalink] New post 13 Jan 2012, 22:21
Thanks Bunuel. Your explanations are awesome.
_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610 :-(

Manager
Manager
avatar
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 178
Location: India
Concentration: Finance, International Business
WE: Information Technology (Investment Banking)
Followers: 3

Kudos [?]: 22 [0], given: 1

Re: Non negative integers in a sequence [#permalink] New post 14 Jan 2012, 02:30
[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.
Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23422
Followers: 3618

Kudos [?]: 28981 [1] , given: 2874

Re: Non negative integers in a sequence [#permalink] New post 14 Jan 2012, 02:48
1
This post received
KUDOS
Expert's post
subhajeet wrote:
Bunuel wrote:
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.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
avatar
Joined: 28 Dec 2010
Posts: 23
Followers: 0

Kudos [?]: 4 [0], given: 2

Re: In a certain sequence, every term after the first is determi [#permalink] New post 08 Feb 2012, 01:33
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 .

Please clarify .
Thanks,
VCG.
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23422
Followers: 3618

Kudos [?]: 28981 [0], given: 2874

Re: In a certain sequence, every term after the first is determi [#permalink] New post 08 Feb 2012, 01:43
Expert's post
verycoolguy33 wrote:
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.

Correct answer: D (63).

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23422
Followers: 3618

Kudos [?]: 28981 [0], given: 2874

Re: In a certain sequence, every term after the first is determi [#permalink] New post 04 Jun 2013, 04:53
Expert's post
Manager
Manager
avatar
Joined: 28 Feb 2012
Posts: 115
Concentration: Strategy, International Business
Schools: INSEAD Jan '13
GPA: 3.9
WE: Marketing (Other)
Followers: 0

Kudos [?]: 21 [0], given: 17

GMAT ToolKit User
Re: In a certain sequence, every term after the first is determi [#permalink] New post 05 Jun 2013, 05:52
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!

Current Student
avatar
Joined: 04 Nov 2012
Posts: 70
Schools: NTU '16 (A)
Followers: 0

Kudos [?]: 23 [0], given: 39

Re: In a certain sequence, every term after the first is determi [#permalink] New post 03 Aug 2013, 21:32
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

so minimum 2 values.

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

Am i right in my reasoning?
Expert Post
1 KUDOS received
Verbal Forum Moderator
Verbal Forum Moderator
User avatar
Joined: 10 Oct 2012
Posts: 628
Followers: 43

Kudos [?]: 599 [1] , given: 135

Premium Member
Re: In a certain sequence, every term after the first is determi [#permalink] New post 03 Aug 2013, 21:47
1
This post received
KUDOS
Expert's post
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.

Hope this helps.
_________________

All that is equal and not-Deep Dive In-equality

Hit and Trial for Integral Solutions

Manager
Manager
User avatar
Joined: 11 Sep 2013
Posts: 79
Followers: 0

Kudos [?]: 15 [0], given: 89

Re: In a certain sequence, every term after the first is determi [#permalink] New post 12 Apr 2014, 10:20
Back-solving should work.
Let 1st = 60, 2nd=120 3rd=240 4th=480 5th=960
Add 2 = 2 4 8 16 32

So, 5th = 960+32=992. 62 numbers+ 0(Zero) = 63 numbers.

Why Zero? Since non-negative numbers, we have to consider Zero as one possible option.
Intern
Intern
avatar
Joined: 28 Dec 2013
Posts: 37
Followers: 0

Kudos [?]: 0 [0], given: 3

Re: In a certain sequence, every term after the first is determi [#permalink] New post 03 Jul 2014, 15:52
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?
Intern
Intern
avatar
Joined: 05 Sep 2012
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: In a certain sequence, every term after the first is determi [#permalink] New post 23 Jul 2014, 11:50
Hi All,

I have a serious doubt regarding the question :

since , ar^4<1000 =>

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.
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23422
Followers: 3618

Kudos [?]: 28981 [0], given: 2874

Re: In a certain sequence, every term after the first is determi [#permalink] New post 24 Jul 2014, 05:31
Expert's post
rhythmboruah wrote:
Hi All,

I have a serious doubt regarding the question :

since , ar^4<1000 =>

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.

Does this make sense?

Check complete solution here: in-a-certain-sequence-every-term-after-the-first-is-determi-126030.html#p1028629
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
User avatar
Joined: 22 Feb 2009
Posts: 229
Followers: 5

Kudos [?]: 49 [0], given: 138

GMAT ToolKit User CAT Tests
Re: In a certain sequence, every term after the first is determi [#permalink] New post 06 Aug 2014, 21:48
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

Intern
Intern
avatar
Joined: 29 Oct 2013
Posts: 20
Followers: 0

Kudos [?]: 6 [0], given: 7

Re: In a certain sequence, every term after the first is determi [#permalink] New post 07 Aug 2014, 09:16
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.

P.S- Please correct me if my approach is wrong!
Re: In a certain sequence, every term after the first is determi   [#permalink] 07 Aug 2014, 09:16
    Similar topics Author Replies Last post
Similar
Topics:
In a sequence of numbers in which each term after the first gmat620 4 14 Nov 2009, 06:38
In a sequence of numbers in which each term after the first Ferihere 3 25 Sep 2007, 05:32
In a sequence of numbers in which each term after the first Jamesk486 2 31 May 2007, 05:42
In a sequence, after the first two terms, each term is the getzgetzu 5 21 Apr 2006, 03:17
In a sequence, after the first two terms, each term is the getzgetzu 1 21 Apr 2006, 03:16
Display posts from previous: Sort by

In a certain sequence, every term after the first is determi

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.