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Bunuel
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Bunuel
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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

Answer B

Please kudo if it helps.
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Didnot get how the answer is C :|

Brunnel can you explain !?
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MSarmah
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.


Answer: C.


Thanks Brunnel !!
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I thought 0 (zero) is divisible by every integer. Therefore I chose B.
Can anyone please correct me?
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lichting
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?
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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.
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Bunuel
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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Thank You Chetan2u, It was much easier than I thought. Great!!!
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