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please see what I wrote above for noting that there is only A as possible answer:
with 1.

if n^2 is multiple of 8 then
n^-2 belongs to [8,16, 24....96]

so
n belongs to [8^2,16^2....96^2]

n belongs also to [0,100}

there is only 8^2 which belongs to both conditions.

with 2 , as explained there are two solutions : 16 or 64. so not enough
hope I explained properly though my choice of A, B,C,D,E was not very clear at first

please see what I wrote above for noting that there is only A as possible answer: with 1.

if n^2 is multiple of 8 then n^-2 belongs to [8,16, 24....96]

so n belongs to [8^2,16^2....96^2]

n belongs also to [0,100}

there is only 8^2 which belongs to both conditions.

with 2 , as explained there are two solutions : 16 or 64. so not enough hope I explained properly though my choice of A, B,C,D,E was not very clear at first

x = 16 can't be a solution for statement 2 as it doesn't satisfy statement 2.
sqrt (16) = 4....and (16)^1/3 = not even an integer

I say B is sufficient. The only number where the square root and the cube root are even integers between 0 and 100 is 64. The reason A is not sufficient is because any number can be a multiple of 8 unless it is specifically states integer multiple...IMHO. _________________