An important note here is that 1 is a factor of all integers (because it evenly divides).

since n > 0 and n^2 is an integer, it indicates that n is an integer.

integer/1 = remainder of 0

1 to the power of whatever is 1, so if n is 1 then 1/1 = remainder 0 and 1^(1/2) = 1/1 = remainder 0.

Now if any other power, we will still have integer/1 = remainder of 0, however if the square root is not an integer then there will be remainder. Hence, the answer is D.

To think of it as a matter of understanding the underlying concept rather than a strategy of picking numbers would follow this logic:

integer/1 = remainder of 0
not integer/1 = some remainder

This is because 1 is a factor of all numbers as mentioned above.

I would prefer to use examples here. For example if n = 4, n^2 is an int and:

res 4/1 = 0

res sqrt(4) = 0

but if n = 3

res n/1 = 0

res sqrt(3) = 0.7777

I think that the approach of: If n^2 is an integer then n is an integer is not correct because we could apply the same for the second condition and say that as n is an integer then sqrt(n) is an integer.

You raise a great point; however, the only case in which the condition you are stating is true is if n = 1. Because every other positive integer in a sqrt will not yield an integer.

I hope this makes sense.

@Salsanousi I think jgastelor is right when he says that n^2 integer doesn't imply n is an integer. You could take n^2 equal every possible integer (like 1,2,3,4,5 to start with) and then n will be equal to +square root of these numbers and not necessarily an integer...

This problem doesn't make sense.
We don't consider remainders for the division of real numbers.
Since we are not sure \(\sqrt{n}\) is an integer, Quantity B, the remainder when \(\sqrt{n}\) is divided by \(1\) doesn't make sense.
A remainder is defined for integers only.

Remember: Remainder is always lesser than divisor.

Quantity A: 0

Quantity B: Value between 0 and 1, that is, 0 <= Quantity B < 1.


Thus we are sure that sometimes A will be greater, and sometimes B, and sometimes both will be equal. Hence relationship can not be determined.

Answer D.

Remember: Remainder is always lesser than divisor.

Quantity A: 0

Quantity B: Value between 0 and 1, that is, 0 <= Quantity B < 1.


Thus we are sure that there is sometimes A will be greater, and sometimes B, and sometimes both will be equal. Hence relationship can not be determined.

Answer D.

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