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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. _________________

excellence is the gradual result of always striving to do better

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

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.

Re: If x and y are positive integers, and x^3y^4 = 2,000, which [#permalink]
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]
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]
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.

Re: If x and y are positive integers and x^3*y^4 = 2000 what is [#permalink]
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

gmatclubot

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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