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

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.

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

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.

[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

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....

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.

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

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

Here's a similar OG question for you to practice:

https://gmatclub.com/forum/if-y-is-the-smallest-positive-integer-such-that-65323.html

Best Regards

Japinder

X^3Y^4=(XY)^3Y=2000=10^3*2; therefore, Y=2 and XY=10--> Answer:D

x^3*y^4=2000
(xy)^3=2000/y=(2/y)*10^3

(xy)=10*(2/y)^(1/3)
Since y is a positive integer taking y=2
D

Hi All,

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)...

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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.

I've done exactly the same

Ive got it right in front of me on my test simulation booklet

But didn't know what to do with the prime factors (just to nervous)

I think if you don't stay calm on test day you'll get nowhere !!

Hi daviddaviddavid,

Most of the questions that you'll face on Test Day will become easier to deal with IF you get in the habit of taking notes (including writing down what the question asks you to solve for).

After reading through this prompt - and taking all of the necessary notes - the last thing you should have written down is.... "(X)(Y) = ?"

In that way, you will know your 'goal' and when you have the necessary information to answer the question that is asked.

GMAT assassins aren't born, they're made,
Rich

Imo D
2^4*5^3=2000 hence xy=10

Sent from my ONE E1003 using GMAT Club Forum mobile app

thank you =)

Ill try my best

at the moment I'm just so disappointed. My test is on Tuesday and I just got 570 today

When I went through the quant questions again, I could solve at least 10 of them (wrong ones) at ease and within seconds but it appears that I'm just freezing under pressure

One questions was x+y = negative ?

(1) x negative
(2) y positive

And I quickly picked C. I guess the algorithm must have thought I'm mentally disabled.


I guess self-confidence and calmness are crucial in order to get a good score.



Sorry for off-topic