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with I. you could have x = 1 and thus y^3=1 and n with a minium value of 2. yeilding an answer of 1. But obviously with larger numbers you could end up with a value over 1000. insuff.

But with II. you know that x>5y and n>x so since they are all positive integers if y=1 then x=5 and y has a minimum value of 6 so you end up with 5^6 which is greater than a thousand. Since this is the minimum case it woudl hold for all positive integers Suff.

Agree with bsmith75.
B
if y = 1, least possible value for x = 6 and n = 7.
So it is (6)^7 > 1000
Based on B, (x/y)^n is going to be greater than 5^6 alyways

1. suff. x=y^3, n>y. Since all are positive integers, the minimum value for x=1 so y=1. (x/y)=1. Although n>y, n could be any positive value and the result of (x/y)^n will aways =1 in this case. Now, if you use x=2, then y=8, creating the fraction, 2/8=1/4. IF you take this vaule to any positive power and since it is a fraction, it will never reach 1000, let alone the value 1. So you can sufficient answer the question as "no".

1. suff. x=y^3, n>y. Since all are positive integers, the minimum value for x=1 so y=1. (x/y)=1. Although n>y, n could be any positive value and the result of (x/y)^n will aways =1 in this case. Now, if you use x=2, then y=8, creating the fraction, 2/8=1/4. IF you take this vaule to any positive power and since it is a fraction, it will never reach 1000, let alone the value 1. So you can sufficient answer the question as "no".

2. suff per bsmith's explanation.

What is the OA?

If x=2, y cannot be 8 from 1). x= y^3 => y = x^1/3 and for x=2, y is not a positive integer.

I agree with Bsmith75. B) it is. _________________