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A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]
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23 Jan 2015, 07:25
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Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]
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23 Jan 2015, 08:13
ans E 243....a5 is int and a6 is not int so a5 should be div by 3 and a6 not therefore 3^5...
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Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]
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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 (an1)/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



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Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]
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24 Jan 2015, 01:54
Here we go
a1 = K
So to be a "cusp" of a sequence
a2 must not be an integer....
So, a2 = 2/3 * ( a1 )
a2 = 2/3 * ( K )
Substitute values for K from the options
A. 3 (integer) B. 16 (not an integer) C. 108 (integer) D. 162 (integer) E. 243 (integer)
Option B is correct



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Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]
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24 Jan 2015, 03:54
mehrdadtaheri92 wrote: 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 (an1)/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
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Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]
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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



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Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]
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24 Jan 2015, 05:45
mehrdadtaheri92 wrote: 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 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
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Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]
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26 Jan 2015, 04:21
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 \(a_n=2\frac{(a_n1)}{3}\) for all n>1, then k could be equal to:
A. 3 B. 16 C. 108 D. 162 E. 243
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION: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.
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Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]
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26 Jan 2015, 04:26
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..
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Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]
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29 Apr 2015, 19:49
Bunnel Great explanation



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A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]
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Updated on: 24 Aug 2017, 10:41
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
Originally posted by Samuelboyle96 on 03 Aug 2017, 11:35.
Last edited by Samuelboyle96 on 24 Aug 2017, 10:41, edited 1 time in total.



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Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]
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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 \(a_n=2\frac{(a_n1)}{3}\) for all n>1, then k could be equal to:
A. 3 B. 16 C. 108 D. 162 E. 243
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION: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 ?



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Re: A term an is called a “cusp” of a sequence if an is an integer but an+ [#permalink]
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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 \(a_n=2\frac{(a_n1)}{3}\) for all n>1, then k could be equal to:
A. 3 B. 16 C. 108 D. 162 E. 243
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION: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.



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



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




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