laiusergiu Wrote:a, x, and k are positive integers. k is odd. Is a divisible by 24?
(1) x^(2a) + 1/[x^(2a)]>2
(2) x^[(k^3)-k] = x^a
What's the source of the question?
Anyway, here goes a solution:
Question is basically asking whether both 3 AND 8 are factors of 'a' or in other words, is 'a' a multiple of 24?
1) For LHS to be greater than 2, x must be greater than 1, since when x=1, LHS = 2, and 2 can not be greater than itself. Thus all positive integer values of x, such that x >1 satisfy this statement. And this makes all positive integer values of 'a' suitable to satisfy the statement.
Thus statement 1 merely tell us that is an integer such that a>0. Statement 1 is NOT SUFFICIENT (to determine whether 'a' is a multiple of 24.)
2)since the two sides of the equation have the same base, we can equate the exponents ===>
(k^3)-k] = a
k[(k^2)-1] = a
k[(k^2)-(1^2)] = a
k(k-1)(k+1) = a
Now, 3 is ALWAYS a divisor/factor of 'a' since 'a' is product of three consecutive (doesn't matter whether positive or negative) integers. So, since k is odd, (k-1)(k)(k+1) is always divisible by 8 since 'a' is zero when k is 1, or 'a' is a multiple of 8 when k is an odd number greater than 1. You can confirm this by plugging in odd integers 3 and 5.
Basically for any positive odd integer k, 'a' is a product of three integers, two of which are different even integers, i.e. k-1 (always even) and k+1 (also always even and different from k-1), hence 'a' is always divisible by 8.
Thus 'a' is divisible by both 3 AND 8, and hence 'a' is a multiple of 24.
when k=3, a= (2)(3)(4)= 24
when k=5, a =(4)(5)
Statement 2 is SUFFICIENT.
Answer is B