Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

51% (01:58) correct
49% (01:58) wrong based on 191 sessions

HideShow timer Statistics

A term an is called a “cusp” of a sequence if \(a_n\) is an integer but \(a_{n+1}\) is not an integer. If \(a_5\) is a cusp of the sequence a1,a2,…,an,… in which \(a_1=k\) and \(a_n=-2\frac{(a_{n-1})}{3}\) for all n>1, then k could be equal to:

Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]

Show Tags

24 Jan 2015, 00:46

THE CORRECT ANSWER IS B NOT E....

WE are told that in a sequence we have first term a1 equal to k and we are given a formula to obtain each of the progression based on that formula

we are given an=-2 (an-1)/3 for n>1 based on a1 is k so we should have a2= -2*k/3

TO obtain a3 , we should replace a2 in the formula and we get a3= -2 *(-2*k/3) /3 so we can get a3=4*k/9

in the same way we can obtain a4 , a4 = -2 * (4k/9)/3 = -8k/27 and for a5 we replace a4 in the formula and get : -2* (-8k/27)/3 = 16k/81

so a5 is : 16k/81 in this point we can see that k is an integer that MUST divisible by 16 and in the given options ONLY OPTION B is the OPTION which is DIVISIBLE BY 16

WE are told that in a sequence we have first term a1 equal to k and we are given a formula to obtain each of the progression based on that formula

we are given an=-2 (an-1)/3 for n>1 based on a1 is k so we should have a2= -2*k/3

TO obtain a3 , we should replace a2 in the formula and we get a3= -2 *(-2*k/3) /3 so we can get a3=4*k/9

in the same way we can obtain a4 , a4 = -2 * (4k/9)/3 = -8k/27 and for a5 we replace a4 in the formula and get : -2* (-8k/27)/3 = 16k/81

so a5 is : 16k/81 in this point we can see that k is an integer that MUST divisible by 16 and in the given options ONLY OPTION B is the OPTION which is DIVISIBLE BY 16

SO ,the correct ans IS B

how can 16k/81 be an integer if k is div by 16.. a5 will be an int if it is div by 81
_________________

Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]

Show Tags

24 Jan 2015, 04:41

Dear Chatan 2 U,

As the problem says a1 is integer so a2 is not so a3 is integer and so on .therefore a5 is integer too.

if we consider X as integer, and placed it infornt of a5 we reach to this equation:

X= 16k/81 and in this case for finding k we change the equation to K= 81*X/16 in this state we must obtain K as integer and as 81 is not divisible by 16 so in order to K to be integer so X must be multiple of 16 and we see here ONLY option B is MULTIPLE of 16 ,SO ans is B

As the problem says a1 is integer so a2 is not so a3 is integer and so on .therefore a5 is integer too.

if we consider X as integer, and placed it infornt of a5 we reach to this equation:

X= 16k/81 and in this case for finding k we change the equation to K= 81*X/16 in this state we must obtain K as integer and as 81 is not divisible by 16 so in order to K to be integer so X must be multiple of 16 and we see here ONLY option B is MULTIPLE of 16 ,SO ans is B

hi, i again feel what he means is that at a(n), the value changes from integer to non integer.. second point, a5 has to be integer so k being integer does not solve the purpose . a5=16*k/81 if a1=k = 16 then a5=16*16/81, which is not integer
_________________

A term an is called a “cusp” of a sequence if \(a_n\) is an integer but \(a_{n+1}\) is not an integer. If \(a_5\) is a cusp of the sequence a1,a2,…,an,… in which \(a_1=k\) and \(a_n=-2\frac{(a_n-1)}{3}\) for all n>1, then k could be equal to:

As with many sequence problems, the wording of this question is designed to be horribly abstract. Let’s just focus on the fact that a5 is an integer but a6 is not an integer.

Putting the sequence into words can help. What do we do to find the next value? We take the previous value and multiply by −2/3. Or even more colloquially, we:

First, flip the sign of the previous one. Second, multiply it by two. Third, divide it by three.

So the second term will be −2/3 times the first one. The third term will be −2/3 times the second one. And so on.

Now, keep in mind that our goal is to find a value of k that will be an integer by the fifth term but not an integer by the sixth term. And this can make your job a lot easier. Positive vs. negative won’t have any bearing on whether a term is an integer or not, and the major operation that will control that will be division. Consider k = 6 as an example (since there are 2s and 3s all over this problem let’s pick a number that’s divisible by both). The first term would be 6, then the second term would be −2/3(6) or -4. Then the third term won’t be an integer, as the “divide by 3” part of the operation will leave you with a decimal.

So when you see that the 5th term of the sequence is going to be:

(−2/3)(−2/3)(−2/3)(−2/3)k, and will be an integer, but that the 6th term of the sequence will be:

(−2/3)(−2/3)(−2/3)(−2/3)(−2/3)k, and will not be an integer, you should see that the real question is “how many times can we evenly divide k by 3?”.

The answer, then, is 4, as it’s that fifth iteration of −2/3 that takes us from “integer” to “noninteger” status.

So kmust contain exactly four 3s, since 81=3^4 and since a5 contains no 3s (recall that a5 was an integer but not a multiple of 3).

Now, factor the answer choices. 3 is already prime factored, but it’s far short of four 3s, so it’s incorrect. 16=2^4; this has no 3s at all, so it’s also incorrect. 108=2^2*3^3; this answer is closer, but it’s still one 3 short, so it’s incorrect too. 162=(2)3^4; that’s the four 3s we need, so this is the correct answer. Finally, 243=3^5; that’s one 3 too many – we needed exactly four – so this is also incorrect.

thanks bunuel.. i was looking for 81 in ans and then settled for next power of 3 ie 243.. how could i miss out 162, a multiple of 81..
_________________

Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]

Show Tags

24 May 2016, 20:08

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]

Show Tags

11 Jun 2017, 07:02

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]

Show Tags

13 Jun 2017, 08:48

1

This post received KUDOS

hi. the original post shows \(a_n=a_n - 1\). but in reality it is \(a_{n-1}\). not sure if my point is coming across as i am new to writing fractions etc on the forum

hi. the original post shows \(a_n=a_n - 1\). but in reality it is \(a_{n-1}\). not sure if my point is coming across as i am new to writing fractions etc on the forum

You are right. Edited the question. Thank you.
_________________

A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]

Show Tags

03 Aug 2017, 11:35

1

This post received KUDOS

I think if you get this right you're on the cusp of greatness. The question isn't too difficult once you understand what they are asking.

So we know A1=K A2 therefore =-2k/3 Because A2=-2(k)/3 A3= 4k/9 because A3 = -2(a2)/3 = -2 ((2k/3)/3) which equals 4k/9 A4 = -8k/27 A5 = 16K/81 A6 = -32K/243

So now we need to look at the solutions. Breaking the numbers into prime factorization is the trick for this part.

1) 3 well we can see that A5 =(16*3)/81 is not an integer 2) 16 is wrong as A5 = (2^4) * (2^4) / (3^4). Well (2^8) /(3^3) is clearly not an integer 3) 108 is wrong as A5 = [(2^4)* (9X13)] / 9^2 . Again we only have one power of 9 on the top and 2 on the bottom. Clearly not an integer 4) 162 = (9^2)* (2^4) / 9^2 this will be an integer

and A6 = -[2^5 * (9^2) *2)] / (9^2 X 3). Notice we have an extra 3 on the bottom therefore this isn't an integer.

5) We can solve E for fun A5= [(2^5) * (9^2)*3] / [(9^2] Clearly an integer A6= 243=(9^2) X 3 . so [(2^5) X (9^2 )*3] / [(9^2) X 3] Clearly an integer

Correct answer D

Last edited by Samuelboyle96 on 24 Aug 2017, 10:41, edited 1 time in total.

Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]

Show Tags

04 Aug 2017, 03:51

Bunuel wrote:

Bunuel wrote:

A term an is called a “cusp” of a sequence if \(a_n\) is an integer but \(a_{n+1}\) is not an integer. If \(a_5\) is a cusp of the sequence a1,a2,…,an,… in which \(a_1=k\) and \(a_n=-2\frac{(a_n-1)}{3}\) for all n>1, then k could be equal to:

As with many sequence problems, the wording of this question is designed to be horribly abstract. Let’s just focus on the fact that a5 is an integer but a6 is not an integer.

Putting the sequence into words can help. What do we do to find the next value? We take the previous value and multiply by −2/3. Or even more colloquially, we:

First, flip the sign of the previous one. Second, multiply it by two. Third, divide it by three.

So the second term will be −2/3 times the first one. The third term will be −2/3 times the second one. And so on.

Now, keep in mind that our goal is to find a value of k that will be an integer by the fifth term but not an integer by the sixth term. And this can make your job a lot easier. Positive vs. negative won’t have any bearing on whether a term is an integer or not, and the major operation that will control that will be division. Consider k = 6 as an example (since there are 2s and 3s all over this problem let’s pick a number that’s divisible by both). The first term would be 6, then the second term would be −2/3(6) or -4. Then the third term won’t be an integer, as the “divide by 3” part of the operation will leave you with a decimal.

So when you see that the 5th term of the sequence is going to be:

(−2/3)(−2/3)(−2/3)(−2/3)k, and will be an integer, but that the 6th term of the sequence will be:

(−2/3)(−2/3)(−2/3)(−2/3)(−2/3)k, and will not be an integer, you should see that the real question is “how many times can we evenly divide k by 3?”.

The answer, then, is 4, as it’s that fifth iteration of −2/3 that takes us from “integer” to “noninteger” status.

So kmust contain exactly four 3s, since 81=3^4 and since a5 contains no 3s (recall that a5 was an integer but not a multiple of 3).

Now, factor the answer choices. 3 is already prime factored, but it’s far short of four 3s, so it’s incorrect. 16=2^4; this has no 3s at all, so it’s also incorrect. 108=2^2*3^3; this answer is closer, but it’s still one 3 short, so it’s incorrect too. 162=(2)3^4; that’s the four 3s we need, so this is the correct answer. Finally, 243=3^5; that’s one 3 too many – we needed exactly four – so this is also incorrect.

Answer: D.

Missed the part when you say "since a5 contains no 3s (recall that a5 was an integer but not a multiple of 3)."

Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]

Show Tags

04 Aug 2017, 06:45

SOUMYAJIT_ wrote:

Bunuel wrote:

Bunuel wrote:

A term an is called a “cusp” of a sequence if \(a_n\) is an integer but \(a_{n+1}\) is not an integer. If \(a_5\) is a cusp of the sequence a1,a2,…,an,… in which \(a_1=k\) and \(a_n=-2\frac{(a_n-1)}{3}\) for all n>1, then k could be equal to:

As with many sequence problems, the wording of this question is designed to be horribly abstract. Let’s just focus on the fact that a5 is an integer but a6 is not an integer.

Putting the sequence into words can help. What do we do to find the next value? We take the previous value and multiply by −2/3. Or even more colloquially, we:

First, flip the sign of the previous one. Second, multiply it by two. Third, divide it by three.

So the second term will be −2/3 times the first one. The third term will be −2/3 times the second one. And so on.

Now, keep in mind that our goal is to find a value of k that will be an integer by the fifth term but not an integer by the sixth term. And this can make your job a lot easier. Positive vs. negative won’t have any bearing on whether a term is an integer or not, and the major operation that will control that will be division. Consider k = 6 as an example (since there are 2s and 3s all over this problem let’s pick a number that’s divisible by both). The first term would be 6, then the second term would be −2/3(6) or -4. Then the third term won’t be an integer, as the “divide by 3” part of the operation will leave you with a decimal.

So when you see that the 5th term of the sequence is going to be:

(−2/3)(−2/3)(−2/3)(−2/3)k, and will be an integer, but that the 6th term of the sequence will be:

(−2/3)(−2/3)(−2/3)(−2/3)(−2/3)k, and will not be an integer, you should see that the real question is “how many times can we evenly divide k by 3?”.

The answer, then, is 4, as it’s that fifth iteration of −2/3 that takes us from “integer” to “noninteger” status.

So kmust contain exactly four 3s, since 81=3^4 and since a5 contains no 3s (recall that a5 was an integer but not a multiple of 3).

Now, factor the answer choices. 3 is already prime factored, but it’s far short of four 3s, so it’s incorrect. 16=2^4; this has no 3s at all, so it’s also incorrect. 108=2^2*3^3; this answer is closer, but it’s still one 3 short, so it’s incorrect too. 162=(2)3^4; that’s the four 3s we need, so this is the correct answer. Finally, 243=3^5; that’s one 3 too many – we needed exactly four – so this is also incorrect.

Answer: D.

Missed the part when you say "since a5 contains no 3s (recall that a5 was an integer but not a multiple of 3)."

why do we say that a5 contains no 3's ?

My solution above is better than Veritas. Veritas explains solutions as if you're a math genius and just missed the answer because you ran out of time. They don't break down the problem thoroughly for those who make mistakes in a number theory. I did above.

Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]

Show Tags

24 Aug 2017, 09:21

Samuelboyle96 wrote:

I think if you get this right you're on the cusp of greatness. The question isn't too difficult once you understand what they are asking.

So we know A1=K A2 therefore =-2k/3 Because A2=-2(k)/3 A3= 4k/9 because A3 = -2(a2)/3 = -2 ((2k/3)/3) which equals 4k/9 A4 = -8k/27 A5 = 16K/81 A6 = -32K/243

So now we need to look at the solutions. Breaking the numbers into prime factorization is the trick for this part.

1) 3 well we can see that A5 =(16*3)/81 is not an integer 2) 16 is wrong as A5 = (2^4) * (2^4) / (3^4). Well (2^8) /(3^3) is clearly not an integer 3) 108 is wrong as A5 = [(2^4)* (9X13)] / 9^2 . Again we only have one power of 9 on the top and 2 on the bottom. Clearly not an integer 4) 162 = (9^2)* (2^4) / 9^2 this will be an integer

and A6 = -[2^5 * (9^2) *2)] / (9^2 X 3). Notice we have an extra 3 on the bottom therefore this isn't an integer.

5) We can solve E for fun A5= [(2^5) * (9^2)*3] / [(9^2)X 3] Clearly an integer A6= 243=(9^2) X 3 . so [(2^5) X (9^2 )*3] / [(9^2) X 3] Clearly an integer

Correct answer D

But this answer is wrong for 5), since you're dividing a5 by 9^2 *3, when you should really just be dividing by 9^2.

Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]

Show Tags

24 Aug 2017, 10:40

brandon7 wrote:

Samuelboyle96 wrote:

I think if you get this right you're on the cusp of greatness. The question isn't too difficult once you understand what they are asking.

So we know A1=K A2 therefore =-2k/3 Because A2=-2(k)/3 A3= 4k/9 because A3 = -2(a2)/3 = -2 ((2k/3)/3) which equals 4k/9 A4 = -8k/27 A5 = 16K/81 A6 = -32K/243

So now we need to look at the solutions. Breaking the numbers into prime factorization is the trick for this part.

1) 3 well we can see that A5 =(16*3)/81 is not an integer 2) 16 is wrong as A5 = (2^4) * (2^4) / (3^4). Well (2^8) /(3^3) is clearly not an integer 3) 108 is wrong as A5 = [(2^4)* (9X13)] / 9^2 . Again we only have one power of 9 on the top and 2 on the bottom. Clearly not an integer 4) 162 = (9^2)* (2^4) / 9^2 this will be an integer

and A6 = -[2^5 * (9^2) *2)] / (9^2 X 3). Notice we have an extra 3 on the bottom therefore this isn't an integer.

5) We can solve E for fun A5= [(2^5) * (9^2)*3] / [(9^2)X 3] Clearly an integer A6= 243=(9^2) X 3 . so [(2^5) X (9^2 )*3] / [(9^2) X 3] Clearly an integer

Correct answer D

But this answer is wrong for 5), since you're dividing a5 by 9^2 *3, when you should really just be dividing by 9^2.

9^2 makes it easier for it to be an integer. That is a typo and I can fix it but the answer is still 100% correct. If 9^2 X 3 is an integer then it's much easier for plain old 9^2 to be an integer.

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...