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# x^3*y^4=2000.What is y? 1. x is an integer 2. y is an

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VP
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x^3*y^4=2000.What is y? 1. x is an integer 2. y is an [#permalink]

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16 Feb 2008, 09:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

x^3*y^4=2000.What is y?
1. x is an integer
2. y is an integer
Director
Joined: 01 May 2007
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16 Feb 2008, 09:44

First rephrase the question...

does (x^3)(y^4) = (5^3)(2^4)?

A tells use x is a integer...we can solve for y?

B tells us y is a integer, we can solve for x?

Is this correct?
VP
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16 Feb 2008, 09:57
jimmyjamesdonkey wrote:

First rephrase the question...

does (x^3)(y^4) = (5^3)(2^4)?

A tells use x is a integer...we can solve for y?

B tells us y is a integer, we can solve for x?

Is this correct?

have a deep look at y. answer is not d
Director
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16 Feb 2008, 10:18
I don't see it...Explanation?
Director
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16 Feb 2008, 10:28
Oh, y could be negative?
Senior Manager
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16 Feb 2008, 10:36
marcodonzelli wrote:
x^3*y^4=2000.What is y?
1. x is an integer
2. y is an integer

(E)

(1) ==> x can be 2, 5 or 10. No way to determine y.
(2) ==> x can be -2, 2 or 5. No way to determine x.

(1) & (2) insufficient
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16 Feb 2008, 14:04
I'm still missing something...can you explain in detail how you can up with 2,5, and 10

and

-2,2, and 5?
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16 Feb 2008, 17:46
giantSwan wrote:
I'm still missing something...can you explain in detail how you can up with 2,5, and 10

and

-2,2, and 5?

Actually my solution was written in a hurry, and was incomplete. Here is a more comprehensive solution.

Statement (1) ==>
(x^3) * (y^4) = 2000, x is an integer
x can be ANY positive integer. You can set any value of x, and there is an equivalent value of y that will result in an equation value of 2000.

Statement (2) ==>
(x^3) * (y^4) = 2000, y is an integer
y can be ANY non-zero integer. You can set any value of y, and there is an equivalent value of x that will result in an equation value of 2000.

Statements (1) & (2) ==>
2000 = (5^3) * (2^4)
You would presume from this breakdown that x=5, and y=2. However, y can be negative (y^4 is always positive). Hence the solution is one of the following:
x=5, y=2; or x=5, y=-2
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Re: Integers and exponents   [#permalink] 16 Feb 2008, 17:46
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