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If x, y, and n are positive integers, is (x/y)^n greater

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If x, y, and n are positive integers, is (x/y)^n greater [#permalink] New post 09 Oct 2005, 16:12
If x, y, and n are positive integers, is (x/y)^n greater than 1,000 ?

(1)x=y^3 and n > y.
(2) x > 5y and n > x.
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 [#permalink] New post 09 Oct 2005, 16:20
I would say (b)

with I. you could have x = 1 and thus y^3=1 and n with a minium value of 2. yeilding an answer of 1. But obviously with larger numbers you could end up with a value over 1000. insuff.

But with II. you know that x>5y and n>x so since they are all positive integers if y=1 then x=5 and y has a minimum value of 6 so you end up with 5^6 which is greater than a thousand. Since this is the minimum case it woudl hold for all positive integers Suff.

B.
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 [#permalink] New post 09 Oct 2005, 17:20
Agree with bsmith75.
B
if y = 1, least possible value for x = 6 and n = 7.
So it is (6)^7 > 1000
Based on B, (x/y)^n is going to be greater than 5^6 alyways
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[#permalink] New post 10 Oct 2005, 06:48
I get D.

1. suff. x=y^3, n>y. Since all are positive integers, the minimum value for x=1 so y=1. (x/y)=1. Although n>y, n could be any positive value and the result of (x/y)^n will aways =1 in this case. Now, if you use x=2, then y=8, creating the fraction, 2/8=1/4. IF you take this vaule to any positive power and since it is a fraction, it will never reach 1000, let alone the value 1. So you can sufficient answer the question as "no".

2. suff per bsmith's explanation.

What is the OA?
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Re: D [#permalink] New post 10 Oct 2005, 09:37
chriswil2005 wrote:
I get D.

1. suff. x=y^3, n>y. Since all are positive integers, the minimum value for x=1 so y=1. (x/y)=1. Although n>y, n could be any positive value and the result of (x/y)^n will aways =1 in this case. Now, if you use x=2, then y=8, creating the fraction, 2/8=1/4. IF you take this vaule to any positive power and since it is a fraction, it will never reach 1000, let alone the value 1. So you can sufficient answer the question as "no".

2. suff per bsmith's explanation.

What is the OA?


If x=2, y cannot be 8 from 1). x= y^3 => y = x^1/3 and for x=2, y is not a positive integer.

I agree with Bsmith75. B) it is.
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 [#permalink] New post 10 Oct 2005, 10:57
B too. Used plugging in a number to find this.

I couldn't figure it out using the formula itself.
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oops [#permalink] New post 11 Oct 2005, 04:39
oops :oops: . Made a mental mistake. Thanks for the heads-up :lol:
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oops   [#permalink] 11 Oct 2005, 04:39
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