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.
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:
Does GMAT RC seem like an uphill battle? e-GMAT 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
a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
20 Dec 2017, 06:07
8
1
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) \(x^{k^3-k} = x^a\)
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.
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) \(x^{k^3-k} = x^a\)
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) = (k-1)*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 k-1 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 k-1 and k+1 are 2 & 4 respectively and their product will have three 2's. Take k=5, then k-1 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 (k-1)*k*(k+1) will have at least one 3 and three 2's , thus will be divisible by 24. Sufficient.
Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
11 Dec 2017, 04:44
1
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 = (K-1)*K* (K+1).
K - 1; K and K+1 is 3 consecutive interger ==> (k-1)K(k+1) is multiple of 3
K is odd ==> K-1 and K+1 must be multiple of 8 ==> A = (k-1)k(K+1) is multiple of 24
Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
21 Dec 2017, 00:07
1
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) \(x^{k^3-k} = x^a\)
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.
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).
Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?
[#permalink]
Show Tags
28 Jan 2018, 19:39
3
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) \(x^{k^3-k} = x^a\)
(1) \(x^{2a} + \frac{1}{x^{2a}}>2\).... Let \(x^{2a}=y\)... \(x^{4a} + 1>2x^{2a}......y^2-2y+1>0.....(y-1)^2>0\).... Since y is POSITIVE integer, y>1, or \(x^{2a}>1\) This means x>1.. Nothing much Insufficient
(2) \(x^{k^3-k} = x^a\)... If x >1... \(x^{k^3-k} = x^a......a=k^3-k=k(k^2-1)=(k-1)k(k+1)\)... 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, \(k^3-k\), take out k common from both terms. So k^3-k=k*k^2-k*1=k(k^2-1)... Now a^2-b^2=(a-b)(a+b).... So k(k-1)(k+1)..
_________________
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.
_________________
close
20% (02:11) correct 
