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

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

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12 Jan 2012, 21:42
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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

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

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13 Jan 2012, 04:17
12
13
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}.

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

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Updated on: 13 Jan 2012, 04:18
10
2
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?

Originally posted by dentobizz on 13 Jan 2012, 03:52.
Last edited by dentobizz on 13 Jan 2012, 04:18, edited 1 time in total.
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Re: Non negative integers in a sequence  [#permalink]

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13 Jan 2012, 02:15
1
2
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
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Re: Non negative integers in a sequence  [#permalink]

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14 Jan 2012, 02:48
1
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}.

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

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03 Aug 2013, 21:47
1
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.
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Re: Non negative integers in a sequence  [#permalink]

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

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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}.

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

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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 .

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

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08 Feb 2012, 01:43
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 .

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.

Hope it's clear.
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Re: In a certain sequence, every term after the first is determi  [#permalink]

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04 Jun 2013, 04:53
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE
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Re: In a certain sequence, every term after the first is determi  [#permalink]

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

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

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

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

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24 Jul 2014, 05:31
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 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
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Re: In a certain sequence, every term after the first is determi  [#permalink]

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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}.

Thanks, I came up with 62, should have read the question more carefully
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Re: In a certain sequence, every term after the first is determi  [#permalink]

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

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18 Jun 2015, 07:22
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.
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Re: In a certain sequence, every term after the first is determi  [#permalink]

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02 Mar 2019, 07:00
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 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}.

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, 07:00
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