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HI Bunuel,

I have a small doubt here....Do positive integers include zero too? If so, we have an undefined value as the answer right? Kinly clarify

Nope .. 0 is neither positive nor negative
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OK so what i remember about zero being a positive integer is wrong...Thanks a ton!
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Bunuel


(2) x>5y and n>x --> \(\frac{x}{y}>5\) also as \(x\), \(y\), and \(n\) are positive integers then the least value of \(x\) is 6 (for \(y=1\)) and the least value of \(n\) is 7 --> so we would have \((# \ more \ than \ 5)^{(at \ least \ 7)}\) which is more than 1,000 (5^7>1,000). Sufficient.

Answer: B.

Can you please explain the 2nd equation again.
I didn;t get this one.
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Bunuel


(2) x>5y and n>x --> \(\frac{x}{y}>5\) also as \(x\), \(y\), and \(n\) are positive integers then the least value of \(x\) is 6 (for \(y=1\)) and the least value of \(n\) is 7 --> so we would have \((# \ more \ than \ 5)^{(at \ least \ 7)}\) which is more than 1,000 (5^7>1,000). Sufficient.

Answer: B.

Can you please explain the 2nd equation again.
I didn;t get this one.

Question: is \((\frac{x}{y})^n>1,00\)?

From (2):

\(x>5y\) --> \(\frac{x}{y}>5\), so \(base=\frac{x}{y}=(# \ more \ than \ 5)\);

\(x>5y\) and \(n>x\) --> as \(x\), \(y\), and \(n\) are positive integers then: the least value \(y\) is 1 --> the least value of \(x\) is 6 (\(x>5=5y_{min}\)) --> the least value of \(n\) is 7 (as \(n>x\));

Is \((\frac{x}{y})^n>1,00\) --> is \((# \ more \ than \ 5)^{(at \ least \ 7)}\)? Answer is YES, as even \(5^7>1,000\).

Hope it's clear.
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Let us substitute numbers to disprove/prove the choices :

We need to predict whether (x/y)^n > 1000

(1) x = y^3 and n > y.

if y = 1 and x = 1 , and n = 2, then it's false.

if y = 10, x = 1000 and n = 1001, it's true

Insufficient

(2) x > 5y and n > x.

Let us take lowest value of y = 1
Then x = 6 at least , and n = 7 at least

So 6^7 > 1000

Another way to look at this is :

x > 5y
=> x/y > 5 and n > 5x => n >= 5 (because these are all positive numbers)

So 5^5 > 1000

Sufficient.

Answer - B
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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agnok
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

Given x,y and n are positive integers

From st 1 we have x= y^3 and n>y so the given expression becomes

(y^2)^n > 1000

now if y = 2 and n = 5 we have 4^5>1000----> yes
but if y=1 and n=5 then we have 1^5>1000-----> no

Not sufficient

St 2 says x>5y and n>x

Let us assume x= 5y so we have 5^n > 1000

now also n> x so if x= 5 then n can be any value integer greater than 5 ----> 5^n>1000 is definitely true
now since x>5y then ----> value of x is more than 5 and since n>x it will always be greater than 1000

Hence ans B
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agnok
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


Excellent Question,,
Here i just plugged in y=1 to calculate the least value of LHS as y increases x increases and so does n hence B is correct
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