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If x and y are positive integers and x^3*y^4 = 2000 what is

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If x and y are positive integers and x^3*y^4 = 2000 what is [#permalink] New post 11 Sep 2008, 13:36
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If x and y are positive integers, and x^3*y^4 = 2,000, which of the following is the value of xy?

A. 2
B. 4
C. 8
D. 10
E. 20
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Re: PS.. More Exp [#permalink] New post 11 Sep 2008, 13:51
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IgnitedMind wrote:
If x and y are positive integers and x^3*y^4 = 2000 what is the value of xy?

2

4

8

10

20


10

(5*2)^{3}*2=2000


D
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If x and y are positive integers, and x^3y^4 = 2,000, which [#permalink] New post 09 Nov 2008, 18:42
If x and y are positive integers, and x^3y^4 = 2,000, which of the following is the value of xy?

A. 2
B. 4
C. 8
D. 10
E. 20
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Re: intergers [#permalink] New post 09 Nov 2008, 20:42
pardon me asking...is the question stem valid as-is? I am finding it hard to deduce a set of +ve integer values for x and y satisfying the below equation

x^3y^4 = 2000

even if you take x=2 and y =1, the lowest possible values, it is still <> 2000.
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Re: intergers [#permalink] New post 10 Nov 2008, 06:22
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2000 = 5^3*2^4
Hence, xy = 10
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Re: intergers [#permalink] New post 10 Nov 2008, 07:13
scthakur wrote:
2000 = 5^3*2^4
Hence, xy = 10


how long did it take you to figure out x and y? I looked at your answer before solving this so just want to estimate which method is better when taking the test.

x^3y^4 = (xy)^3 * y = 2000

a. if xy = 2 => 2^3 = 8, can't find y
b. xy = 4 (same as above)
c. xy = 8, 8^3 = 512; 2000/512 is not integer
d. xy = 10, 10^3 = 1000, y = 2000/1000 = 2
e. xy = 20, 20^3 = 8000 > 2000

so the answer is D
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Re: intergers [#permalink] New post 10 Nov 2008, 09:15
Actually, I had seen this problem before. Haaving said that, in such types of problems, it is better to break the big number into squares, cubes, etc. of smaller numbers.
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Re: intergers [#permalink] New post 10 Nov 2008, 10:49
[quote="Jcpenny"]If x and y are positive integers, and x^3y^4 = 2,000, which of the following is the value of xy?

A. 2
B. 4
C. 8
D. 10
E. 20

1000*2 = 10^3*2 = 5^3*2^4 thus xy = 10...D
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Re: intergers [#permalink] New post 11 Nov 2008, 07:28
hit n trial can help here.
we can easily guess that either x or y must be 5 here.

if y=5 we cant divide 2000 by 625 ...so x=5...and there u go...

alternatively, we can think of the possible values of (xy)3 ...where (xy)3 =1000 is most plausible option considering y to be an integer....
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Re: If x and y are positive integers, and x^3y^4 = 2,000, which [#permalink] New post 17 Mar 2014, 16:54
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Re: If x and y are positive integers and x^3*y^4 = 2000 what is [#permalink] New post 17 Mar 2014, 17:01
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Re: If x and y are positive integers and x^3*y^4 = 2000 what is [#permalink] New post 08 Apr 2014, 18:52
for this problem, i factored 2000 --> 2, 2, 2, 2, 5, 5, 5... or (5*3)(2*4).

based on the factoring, x = 5 and y = 2... so xy = (5)(2) = 10

the correct answer is d.
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Re: If x and y are positive integers and x^3*y^4 = 2000 what is [#permalink] New post 08 Apr 2014, 22:52
In such question it is best to see the options available.
x^3*y^4 = 2000 is essentially same as (xy)^3*y = 2000
Now, if you look at the available choices, it is 2, 4, 8, 10, 20
You can figure out looking at choices that 2, 4, 8 are too small to satisfy the equation while 20 is too large.
Hence, lets check with 10. If xy = 10, xy^3 = 1000, so y = 2000/1000 = 2 and x = 10/2 = 5.
5 & 2 being positive integers, xy=10 is the correct answer.
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Re: If x and y are positive integers and x^3*y^4 = 2000 what is [#permalink] New post 11 Apr 2014, 06:57
My Analysis with Strategy..
1 Positive Integers ( + ) , 2 Powers , 3 Prime factorization
So Prime factorization of 2000 write in Power form = 2^4 5^3
So x = 5 , y = 2 ; 5 * 2 = 10 :)
Re: If x and y are positive integers and x^3*y^4 = 2000 what is   [#permalink] 11 Apr 2014, 06:57
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