Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 18 08:00 AM PDT  09:00 AM PDT Learn an intuitive, systematic approach that will maximize your success on Fillintheblank GMAT CR Questions. Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss!
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58453

a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
11 Dec 2017, 00:50
Question Stats:
20% (02:11) correct 80% (02:40) wrong based on 140 sessions
HideShow timer Statistics
a, x, and k are positive integers. k is odd. Is a divisible by 24? (1) \(x^{2a} + \frac{1}{x^{2a}}>2\) (2) \(x^{k^3k} = x^a\)
Official Answer and Stats are available only to registered users. Register/ Login.
_________________




Math Expert
Joined: 02 Sep 2009
Posts: 58453

a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
20 Dec 2017, 06:07
MSarmah wrote: Didnot get how the answer is C Brunnel can you explain !? This is a hard problem and the solutions above are not correct. a, x, and k are positive integers. k is odd. Is a divisible by 24?(1) \(x^{2a} + \frac{1}{x^{2a}}>2\). Notice that since x and a are positive integers, the above inequality holds true for any values except when x = 1. If x = 1, then \(x^{2a} + \frac{1}{x^{2a}} = 1 + 1 = 2\) If x > 1, then \(x^{2a} + \frac{1}{x^{2a}} = (integer \ greater \ than \ 2) + (positive \ number \ less \ than \ 1) >2\) So, this statement tells us that x ≠ 1. Not sufficient. (2) \(x^{k^3k} = x^a\) If x = 1, then a and k can be any integers, because 1^anything = 1, no matter what k^3  k and a are. If x ≠ 1, then we can equate the powers: \(a = k^3  k = k(k^2  1)= (k  1)k(k + 1)\). So, a is a product of three consecutive integers, thus one of them is a multiple of 3. Also, since given that k is odd, then (k  1) and (k + 1) are consecutive even integer, which imply that one of them is a multiple of 4, so a is a multiple of 2*4 = 8. Thus, in this case a becomes a multiple of both 3 and 8, so a multiple of 3*8 = 24. Not sufficient. (1)+(2) From (1) x ≠ 1, thus from (2) a multiple of 3*8 = 24. Sufficient. Answer: C.
_________________




Retired Moderator
Joined: 22 Aug 2013
Posts: 1428
Location: India

Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
11 Dec 2017, 04:20
Bunuel wrote: a, x, and k are positive integers. k is odd. Is a divisible by 24?
(1) \(x^{2a} + \frac{1}{x^{2a}}>2\)
(2) \(x^{k^3k} = x^a\) 24 = 2^3 * 3 So for any integer to be divisible by 24, that number must have at least three 2's and one 3 in its prime factorisation. (1) If x is any positive integer apart from 1, then x^2a + 1/x^2a will ALWAYS be greater than 2 (given that 'a' is also a positive integer). For x=1, x^2a + 1/x^2a will be 1^2a + 1/x^2a = 1+1=2. But for x=2, x^2a + 1/x^2a= 2^2a + 1/2^2a. No matter which value of 'a' (positive integer) we put here, it will come out to be > 2. And with increasing values of x and 'a', the value of x^2a + 1/x^2a will keep on increasing further, and always > 2. So we cant say whether a is divisible by 24 or not. Insufficient. (2) From this statement we can conclude that a = k^3  k = k (k^2  1) = (k1)*k*(k+1). Now 'a' is a product of three consecutive integers thus it will be divisible by 3. Also given that k is odd, so both k+1 and k1 will be even, and also, one of them will also be a multiple of 4. So this guarantees three 2's also. Eg, take k=3, then k1 and k+1 are 2 & 4 respectively and their product will have three 2's. Take k=5, then k1 and k+1 will be 4 and 6 respectively and their product will also have three 2's.. So for any odd integer value of k, the product (k1)*k*(k+1) will have at least one 3 and three 2's , thus will be divisible by 24. Sufficient. Hence B answer



Manager
Joined: 25 Jul 2017
Posts: 55

Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
11 Dec 2017, 04:44
From 1:
X^ (2a) = (X^a)^2 ==> positive number 1/ (X^2a) = (1/x^a)^2 ==> positive number ==> X^ (2a) + 1/ (X^2a) = (X^a)^2 + (1/x^a)^2 is always greater than 2, no matter which value of a (1) is sufficient
From 2 ==> K^3  K = a = (K1)*K* (K+1).
K  1; K and K+1 is 3 consecutive interger ==> (k1)K(k+1) is multiple of 3
K is odd ==> K1 and K+1 must be multiple of 8 ==> A = (k1)k(K+1) is multiple of 24
Answer B
Please kudo if it helps.



Intern
Joined: 19 Aug 2017
Posts: 15

Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
20 Dec 2017, 04:03
Didnot get how the answer is C Brunnel can you explain !?



Intern
Joined: 19 Aug 2017
Posts: 15

Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
21 Dec 2017, 00:07
Bunuel wrote: MSarmah wrote: Didnot get how the answer is C Brunnel can you explain !? This is a hard problem and the solutions above are not correct. a, x, and k are positive integers. k is odd. Is a divisible by 24?(1) \(x^{2a} + \frac{1}{x^{2a}}>2\). Notice that since x and a are positive integers, the above inequality holds true for any values except when x = 1. If x = 1, then \(x^{2a} + \frac{1}{x^{2a}} = 1 + 1 = 2\) If x > 1, then \(x^{2a} + \frac{1}{x^{2a}} = (integer \ greater \ than \ 2) + (positive \ number \ less \ than \ 1) >2\) So, this statement tells us that x ≠ 1. Not sufficient. (2) \(x^{k^3k} = x^a\) If x = 1, then a and k can be any integers, because 1^anything = 1, no matter what k^3  k and a are. If x ≠ 1, then we can equate the powers: \(a = k^3  k = k(k^2  1)= (k  1)k(k + 1)\). So, a is a product of three consecutive integers, thus one of them is a multiple of 3. Also, since given that k is odd, then (k  1) and (k + 1) are consecutive even integer, which imply that one of them is a multiple of 4, so a is a multiple of 2*4 = 8. Thus, in this case a becomes a multiple of both 3 and 8, so a multiple of 3*8 = 24. Not sufficient. (1)+(2) From (1) x ≠ 1, thus from (2) a multiple of 3*8 = 24. Sufficient. Answer: C. Thanks Brunnel !!



Intern
Joined: 17 Sep 2017
Posts: 45

Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
21 Dec 2017, 07:53
I thought 0 (zero) is divisible by every integer. Therefore I chose B. Can anyone please correct me?



Math Expert
Joined: 02 Sep 2009
Posts: 58453

Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
21 Dec 2017, 07:57
lichting wrote: I thought 0 (zero) is divisible by every integer. Therefore I chose B. Can anyone please correct me? Yes, 0 is divisible by every integer, except 0 itself but what this has to do with the question? How did you get that a = 0?
_________________



Intern
Joined: 30 Sep 2016
Posts: 5

Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
28 Jan 2018, 19:19
Hello Bunuel can you please explain me how did you get this (k−1)k(k+1)?. I don't know if I'm missing a step or rule on this one. Although it is straightforward how you got to a=k3−k=k(k2−1).
Thank you.



Math Expert
Joined: 02 Aug 2009
Posts: 7978

Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
28 Jan 2018, 19:39
Bunuel wrote: a, x, and k are positive integers. k is odd. Is a divisible by 24?
(1) \(x^{2a} + \frac{1}{x^{2a}}>2\)
(2) \(x^{k^3k} = x^a\) (1) \(x^{2a} + \frac{1}{x^{2a}}>2\).... Let \(x^{2a}=y\)... \(x^{4a} + 1>2x^{2a}......y^22y+1>0.....(y1)^2>0\).... Since y is POSITIVE integer, y>1, or \(x^{2a}>1\) This means x>1.. Nothing much Insufficient (2) \(x^{k^3k} = x^a\)... If x >1... \(x^{k^3k} = x^a......a=k^3k=k(k^21)=(k1)k(k+1)\)... Sum of three CONSECUTIVE integers where k is odd... So one of the digits will be MULTIPLE of 3, one of them will be ATLEAST div by 2 and other even number by ATLEAST 4, so a is div by 2*3*4=24..yes If x=1 K and a can be any values as 1^anything is 1 Insufficient Combined x>1, so a is div by 24 as shown from statement II Suff.. Tavo, \(k^3k\), take out k common from both terms. So k^3k=k*k^2k*1=k(k^21)... Now a^2b^2=(ab)(a+b).... So k(k1)(k+1)..
_________________



Intern
Joined: 30 Sep 2016
Posts: 5

Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
29 Jan 2018, 13:48
Thank You Chetan2u, It was much easier than I thought. Great!!!



NonHuman User
Joined: 09 Sep 2013
Posts: 13259

Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
09 Jul 2019, 11:05
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, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
09 Jul 2019, 11:05






