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a, x, and k are positive integers. k is odd. Is a divisible by 24?
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20 Dec 2017, 06:07
5
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?
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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?
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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?
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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) \(x^{k^3-k} = x^a\)
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?
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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?
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28 Jan 2018, 19:39
2
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)..
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