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
Sequence
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- ssmiles08
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I got C as my answer.
I took the n as my first term:
n
n*a
n*a^2
n*a^3
n*a^4--->5th term
Since a has to be an integer greater than 1, lets chose 2 as a to maximize the value of n
2^4 = 16
1000/16 = 125/2 = 62.5
the greatest integer less than 62.5 is 62.
I took the n as my first term:
n
n*a
n*a^2
n*a^3
n*a^4--->5th term
Since a has to be an integer greater than 1, lets chose 2 as a to maximize the value of n
2^4 = 16
1000/16 = 125/2 = 62.5
the greatest integer less than 62.5 is 62.
- vineetbatra
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- Stuart@KaplanGMAT
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So close and yet so far!vineetbatra wrote:This is a GP question.
Formula is a*r^(n-1), where a is the first number, r is the constant and n is the number of times constant is applied
I agree that the the answer should be 62, becuase with 63 the answer will be 1008.
0 is a non-negative integer... so the possible values are {0, 1, 2, ... 62}, for a total of 63 possible values.
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- vineetbatra
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[quote="Stuart Kovinsky"][quote="vineetbatra"]This is a GP question.
Formula is a*r^(n-1), where a is the first number, r is the constant and n is the number of times constant is applied
I agree that the the answer should be 62, becuase with 63 the answer will be 1008.[/quote]
So close and yet so far!
0 is a non-negative integer... so the possible values are {0, 1, 2, ... 62}, for a total of 63 possible values.[/quote]
Hi Stuart
thanks for your help and support.
Could you please run through the above in more detail, I'm missing something.
Thanks
Ian
Formula is a*r^(n-1), where a is the first number, r is the constant and n is the number of times constant is applied
I agree that the the answer should be 62, becuase with 63 the answer will be 1008.[/quote]
So close and yet so far!
0 is a non-negative integer... so the possible values are {0, 1, 2, ... 62}, for a total of 63 possible values.[/quote]
Hi Stuart
thanks for your help and support.
Could you please run through the above in more detail, I'm missing something.
Thanks
Ian
- Stuart@KaplanGMAT
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I'm actually not a big fan of memorizing hundreds and hundreds of formulas for test day - that's definitely not the approach I'd have taken to this question.drian wrote:Hi StuartStuart Kovinsky wrote:So close and yet so far!vineetbatra wrote:This is a GP question.
Formula is a*r^(n-1), where a is the first number, r is the constant and n is the number of times constant is applied
I agree that the the answer should be 62, becuase with 63 the answer will be 1008.
0 is a non-negative integer... so the possible values are {0, 1, 2, ... 62}, for a total of 63 possible values.
thanks for your help and support.
Could you please run through the above in more detail, I'm missing something.
Thanks
Ian
Let's start by reexamining the question:
Step 1 of the Kaplan method for problem solving: study the question stem and the answers.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
Here we see that we've got a complicated word problem about sequences and answer choices that are numbers. To solve, we're going to need to understand the sequence.
We see that we're multiplying each term by "an integer constant greater than 1" to form the next term. We also see that the question asks about non-negative first terms. We think: "non-negative" means "0 or positive".
Finally, we see that the 5th term must be less than 1000.
Step 2 of the Kaplan method for PS: identify the exact question.
We want to know the maximum number of non-negative integers that could be the first term of the sequence.
Step 3 of the Kaplan method for PS: choose the most efficient approach.
As with most number property questions, we can either use principles or pick numbers. On this question, picking numbers will almost certainly be the quickest way to proceed.
Since we want to maximize the number of possible starting terms, we want to make our multiplier as small as possible. In this question, we let our multiplier = 2.
If the 1st term is x, the 5th term is x * 2 * 2 * 2 * 2 = 16x.
We know that 16x < 1000. To find the biggest possible value for x, let's divide both sides by 16:
x < 1000/16
and a bit of long division tells us that:
x < 62.something
(we couldn't care less what comes after the decimal point).
Therefore, the biggest possible integer value for x is 62.
At this point you mght be tempted to select 62 as your answer. Questions on the GMAT are often constructed to punish people who don't read the question carefully, which is exactly why there's a step 4 of the Kaplan method!
Step 4 of the Kaplan method for PS: double check the question.
We want the maximum n umber of non-negative 1st temrs, which means we can't forget about 0! Therefore, our set of possible 1st terms is:
{0, 1, 2, 3, ... 62}
which contains 63 possibilities: choose (D).
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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" every term after the first is determined by multiplying the previous term by an integer constant greater than 1"
so if the 1st term is 0 and say the constant is 2 then we have:
1st no = 0
2nd no = 0*2 = 0
3rd no = 0*2 = 0
and so on..
thus, its just forms a set of 0s
can this be regarded as a sequence??
if not, then the answere is 62!
so if the 1st term is 0 and say the constant is 2 then we have:
1st no = 0
2nd no = 0*2 = 0
3rd no = 0*2 = 0
and so on..
thus, its just forms a set of 0s
can this be regarded as a sequence??
if not, then the answere is 62!
- Stuart@KaplanGMAT
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Yes - see my post above, 0 needs to be included on the list of "non-negative integers".meet.anup wrote:" every term after the first is determined by multiplying the previous term by an integer constant greater than 1"
so if the 1st term is 0 and say the constant is 2 then we have:
1st no = 0
2nd no = 0*2 = 0
3rd no = 0*2 = 0
and so on..
thus, its just forms a set of 0s
can this be regarded as a sequence??
if not, then the answere is 62!
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course