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.
Answer: C.