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So I am surmizing, that to get the answer for this

1) Do a prime factorization of 2000. 2 raised to 4 and 5 raised to 3

2) Realize that you have a cube and power of four in the multiplication. The integer raised to the power of four can either be a negative or a positive.

Now, (-2)^4 = (2)^4= 16... thats a good catch here, + 2 & -2 raised to a +ve exponent yield the same positive result. Hence, y does not have definite value, it could be either +2 or -2.

St1: x is an integer - insufficient, coz value of y can be anything, right from an integer to a fractional value...( the only fact that is derived here is x will be definitely positive) St2: y is an integer - insufficient, x can take any value.

St1 & St2 - x should be positive, y can be positive or negative..

came up with one solution using (1) 2000 = 1000 * 2 = 10^3 * (sqrt(sqrt(2)) ) ^ 4

similar solution can be obtained for (2)

however, many questions remain unanswered. or simply put, if you plug in multiple values, you may or may not get 2000 as the product. _________________

The catch point here is to pick numbers of different varieties in order to try to deny and support a statement.. Since both are integers, each can be +ve or -ve.

What is the fastest way to determine that x and y are 5 and +/- 2?

Or would an easier way to answer this question be simply to note that because y is raised to an even exponent it will have to have an +/- answer and therefore the answer will be E? _________________

To those who have struggled with them, the mountains reveal beauties that they will not disclose to those who make no effort. That is the reward the mountains give to effort. And it is because they have so much to give and give it so lavishly to those who will wrestle with them that men love the mountains and go back to them again and again. The mountains reserve their choice gifts for those who stand upon their summits. (Sir Francis Younghusband)

In the DS section -- its useful to take a moment to just think about what concept the GMAT is likely trying to test. This makes it much easier to solve the problem than just using the brute force method.

Straight E. The key point is to remember that even exponents mask the sign of the base number. The rest of the solution is as illustriously noted by other posters on this thread.

Each statement alone is insufficient. When considering together we have: \(x^3*y^4 = 5^3*2^4=2000\), but since \(y\) is in even power it can be 2 as well as -2. Not sufficient.

Hi Bunuel, I have a question here, so when you considered both the statement and got x^3*y^4 = 5^3*2^4=2000. i want to know how you reached there? did you try plugging in x=1 ,2, 3 , 4 and 5 and y =1 , 2 etc?just trying to understand the thought process . when i tried to take both the statements i could not do that .

Each statement alone is insufficient. When considering together we have: \(x^3*y^4 = 5^3*2^4=2000\), but since \(y\) is in even power it can be 2 as well as -2. Not sufficient.

Answer: E.

_________________

--------------------------------------------------------------------------------------------- Kindly press +1 Kudos if my post helped you in any way