Sequence

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 238
Joined: Tue Feb 10, 2009 8:44 am
Thanked: 9 times

Sequence

by avenus » Sun May 31, 2009 10:45 am
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

User avatar
Master | Next Rank: 500 Posts
Posts: 472
Joined: Sun Mar 29, 2009 6:54 pm
Thanked: 56 times

by ssmiles08 » Sun May 31, 2009 10:52 am
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.

Master | Next Rank: 500 Posts
Posts: 238
Joined: Tue Feb 10, 2009 8:44 am
Thanked: 9 times

by avenus » Tue Jun 02, 2009 1:10 am
OA is D

User avatar
Master | Next Rank: 500 Posts
Posts: 355
Joined: Thu Feb 19, 2009 12:42 pm
Thanked: 2 times
Followed by:1 members

Gp

by vineetbatra » Tue Jun 02, 2009 6:36 am
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.

User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

Re: Gp

by Stuart@KaplanGMAT » Tue Jun 02, 2009 12:13 pm
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.
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.
Image

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

User avatar
Master | Next Rank: 500 Posts
Posts: 355
Joined: Thu Feb 19, 2009 12:42 pm
Thanked: 2 times
Followed by:1 members

Bugger

by vineetbatra » Tue Jun 02, 2009 12:55 pm
oh bugger, devil is in the detail.

thanks for the clarification.

User avatar
Master | Next Rank: 500 Posts
Posts: 472
Joined: Sun Mar 29, 2009 6:54 pm
Thanked: 56 times

by ssmiles08 » Tue Jun 02, 2009 2:44 pm
Oh grr totally missed that!! :)

Senior | Next Rank: 100 Posts
Posts: 40
Joined: Sun May 31, 2009 10:17 pm
Thanked: 1 times

Need some more info for this problem

by eustudent » Tue Jun 02, 2009 8:33 pm
can you please privude us, the readers, with some more clarigication how do you apply the GS formula in this particular problem; and how do you get to the final decision?

Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Sat May 30, 2009 8:33 pm

Re: Gp

by drian » Tue Jun 02, 2009 8:56 pm
[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

Newbie | Next Rank: 10 Posts
Posts: 7
Joined: Tue Jun 02, 2009 10:34 am
Thanked: 1 times

by vpmba2009 » Wed Jun 03, 2009 9:51 am
I still don't get it. Any other way to explain to easily understand this?
Thanks!

User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

Re: Gp

by Stuart@KaplanGMAT » Wed Jun 03, 2009 10:41 am
drian wrote:
Stuart Kovinsky wrote:
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.
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.
Hi Stuart

thanks for your help and support.
Could you please run through the above in more detail, I'm missing something.

Thanks

Ian
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.

Let's start by reexamining the question:
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
Step 1 of the Kaplan method for problem solving: study the question stem and the answers.

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).
Image

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

Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Thu Jun 05, 2008 11:25 am

by meet.anup » Thu Jul 09, 2009 3:42 am
" 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!

User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Thu Jul 09, 2009 7:34 am
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!
Yes - see my post above, 0 needs to be included on the list of "non-negative integers".
Image

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