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11 Dec 2017, 00:50
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
26% (02:28) correct
74%
(02:46)
wrong
based on 312
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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) \(x^{2a} + \frac{1}{x^{2a}}>2\)
(2) \(x^{k^3-k} = x^a\)
20 Dec 2017, 06:07
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MSarmah
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.
General Discussion
11 Dec 2017, 04:20
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Bunuel
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.
Hence B answer
