Author 
Message 
TAGS:

Hide Tags

Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 439
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

In a certain sequence, every term after the first is determi
[#permalink]
Show Tags
12 Jan 2012, 22:42
Question Stats:
26% (02:35) correct 74% (02:42) wrong based on 732 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 nonnegative integer values possible for the first term? A) 60 B) 61 C) 62 D) 63 E) 64
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730




Math Expert
Joined: 02 Sep 2009
Posts: 60594

Re: Non negative integers in a sequence
[#permalink]
Show Tags
13 Jan 2012, 05:17
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 nonnegative 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 nonnegative 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 nonnegative 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.
_________________




Retired Moderator
Joined: 23 Jul 2010
Posts: 421
GPA: 3.4
WE: General Management (NonProfit and Government)

Re: Non negative integers in a sequence
[#permalink]
Show Tags
Updated on: 13 Jan 2012, 05:18
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 nonnegative 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 nonnegative 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 6064. 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 nonnegative 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?
Originally posted by dentobizz on 13 Jan 2012, 04:52.
Last edited by dentobizz on 13 Jan 2012, 05:18, edited 1 time in total.




Manager
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 119
Location: India
Concentration: Finance, International Business
WE: Information Technology (Investment Banking)

Re: Non negative integers in a sequence
[#permalink]
Show Tags
13 Jan 2012, 03:15
Ans is C
From the statement we can make an eq as a_n = a_(n1)*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



Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 439
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

Re: Non negative integers in a sequence
[#permalink]
Show Tags
13 Jan 2012, 23:21
Thanks Bunuel. Your explanations are awesome.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Manager
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 119
Location: India
Concentration: Finance, International Business
WE: Information Technology (Investment Banking)

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 nonnegative 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.



Math Expert
Joined: 02 Sep 2009
Posts: 60594

Re: Non negative integers in a sequence
[#permalink]
Show Tags
14 Jan 2012, 03:48
subhajeet wrote: Bunuel wrote: Thus, \(x*2^4<1,000\) >\(x<\frac{1,000}{16}=62,5\) > as the first term must be a nonnegative 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 nonnegative 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.
_________________



Intern
Joined: 28 Dec 2010
Posts: 18

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 .
Please clarify . Thanks, VCG.



Math Expert
Joined: 02 Sep 2009
Posts: 60594

Re: In a certain sequence, every term after the first is determi
[#permalink]
Show Tags
08 Feb 2012, 02:43
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.
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 60594

Re: In a certain sequence, every term after the first is determi
[#permalink]
Show Tags
04 Jun 2013, 05:53
Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE
_________________



Manager
Joined: 28 Feb 2012
Posts: 103
Concentration: Strategy, International Business
GPA: 3.9
WE: Marketing (Other)

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 nonnegative 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.



Intern
Joined: 04 Nov 2012
Posts: 47

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
so minimum 2 values.
we cannot take r =6 because 6^4>1000
Am i right in my reasoning?



Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 576

Re: In a certain sequence, every term after the first is determi
[#permalink]
Show Tags
03 Aug 2013, 22:47
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 NONNEGATIVE 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.
_________________



Manager
Joined: 11 Sep 2013
Posts: 131
Concentration: Finance, Finance

Re: In a certain sequence, every term after the first is determi
[#permalink]
Show Tags
12 Apr 2014, 11:20
Backsolving 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 nonnegative numbers, we have to consider Zero as one possible option.



Intern
Joined: 05 Sep 2012
Posts: 1

Re: In a certain sequence, every term after the first is determi
[#permalink]
Show Tags
23 Jul 2014, 12:50
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.



Math Expert
Joined: 02 Sep 2009
Posts: 60594

Re: In a certain sequence, every term after the first is determi
[#permalink]
Show Tags
24 Jul 2014, 06:31
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 nonnegative 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 nonnegative 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: inacertainsequenceeverytermafterthefirstisdetermi126030.html#p1028629
_________________



Manager
Joined: 22 Feb 2009
Posts: 151

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 nonnegative 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 nonnegative 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 nonnegative 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



Intern
Joined: 29 Oct 2013
Posts: 18

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.e2.
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!



Manager
Joined: 10 Jun 2015
Posts: 110

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 nonnegative 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.



Senior Manager
Joined: 23 Nov 2016
Posts: 325
GMAT 1: 690 Q50 V33

Re: In a certain sequence, every term after the first is determi
[#permalink]
Show Tags
02 Mar 2019, 08:00
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 nonnegative 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 nonnegative 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 nonnegative 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. How 0 is possible? If it is 0 then every successive term will become 0 as we are multiplying with constant




Re: In a certain sequence, every term after the first is determi
[#permalink]
02 Mar 2019, 08:00






