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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.
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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]

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17 Mar 2014, 17:54

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

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

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08 Apr 2014, 23:52

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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]

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11 Apr 2014, 07:57

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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

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In questions like this one, it's a good idea to start by Prime Factorizationof the big number. Some of the students who posted their solutions above have done this already.

This question involves some Number Properties; if you recognize them, think logically and choose the SIMPLEST examples possible, then you'll get to the correct answer relatively quickly.

We're told that (X^3)(Y^4) = 2,000

Since we're dealing with a product that ends in 0, at least one of the variables is a multiple of 5 and at least one of the variables is even.

So....choose two different numbers....the most obvious multiple of 5 and the most obvious even....

You'll end up with 5 and 2....Plug those in (one for X and one for Y) - it'll take no more than 2 attempts to find the one that 'fits' and you'll have the value of (X)(Y)...

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

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02 Jun 2015, 04:03

D 2 and 5 are in equal numbers in 1000 i.e. 2^3 and 5^3 and thus in 2 x 1000 --- number of 2 = 1+no. of 5's. Thus there product would be 20.
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