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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
A) 60
B) 61
C) 62
D) 63
E) 64
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
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.
_________________
A. 60
B. 61
C. 62
D. 63
E. 64
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.
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enigma123
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?
General Discussion
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
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
[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.[/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.
subhajeet
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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Posts: 17
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 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.
verycoolguy33
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.
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Schools: NTU '16 (A)
Posts: 42
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?
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?
12bhang
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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Posts: 1
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.
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.
rhythmboruah
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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Bunuel
Thanks, I came up with 62, should have read the question more carefully
how do you conclude that each time we are multiplying the previous term with an integer, we are multiplying with the same integer every time?
Bunuel
ashKing12
Because “integer constant” means one fixed multiplier is used for every term in the sequence.
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