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505-555 Level|   Algebra|                           
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Bunuel
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Bunuel
If x is a positive integer, what is the value of x ?

(1) \(x^2=\sqrt{x}\)

(2) \(\frac{n}{x} = n\) and \(n\neq 0\)


Another approach for those that they are quite confident with exponents:

1)Squaring both sides of X^2 = sqrt(X) you get X^4 = X. Now divide both sides with X and you get X^3 = 1 (You can divide by x as x>0). Now the tricky thing comes in...you have to raise both sides to 1/3 power in order to get

(X^3) ^ 1/3 = 1^(1/3) which give X = 1^(1/3). The positive 1 is equal to 1 if is raised to any positive power which is the case. Hence Sufficient.

2) VERY Straightforward
Multiply both sides by X so you can get n = x*n and then divide both sides by n in order to get n/n = (n*x) / n which will give x =1. Hence Sufficient.


I am not sure why I went one step further with statement (1). I am quite suprised that all of you did the x(x^3-1) = 0 ... Can you please explain why you stopped there?
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Bunuel
If x is a positive integer, what is the value of x ?

(1) \(x^2=\sqrt{x}\)

(2) \(\frac{n}{x} = n\) and \(n\neq 0\)


(DS08306)

First we square both sides of statement 1)

\(x^4 = x\) then it becomes \(x^4 - x = 0\) and taking x as a common factor it becomes \(x (x^3 - 1) = 0\) from the question stem we know x is a positive integer so it cannot be 0

\(x = 1\) sufficient.

Eliminate B/C/E

check statement 2)

\(\frac{n}{x} = n\)

\(xn = n\) then \(xn - n = 0\) then \(n (x-1) = 0\)we know n cannot be zero so x = 1 sufficient.

Answer choice D
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Bunuel
If x is a positive integer, what is the value of x ?

(1) \(x^2=\sqrt{x}\)

(2) \(\frac{n}{x} = n\) and \(n\neq 0\)


(DS08306)
Solution:

Question Stem Analysis:


We need to determine the value of x, given that it is a positive integer.

Statement One Alone:

The only positive integer (and actually the only positive number) whose square is equal to its square root is 1. Therefore x = 1. Statement one alone is sufficient.

Statement Two Alone:

Multiplying both sides of the equation by x (since it is not 0), we have:

n = nx

Since n is also not 0, dividing both sides by n, we have:

1 = x

Statement two alone is sufficient.

Answer: D
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Bunuel
If x is a positive integer, what is the value of x ?

(1) \(x^2=\sqrt{x}\)

(2) \(\frac{n}{x} = n\) and \(n\neq 0\)


x is a positive integer (given)

From statement 1
\(x^2=\sqrt{x}\)
squaring and re arranging we get x^4-x=0, x=0,1
As x is a positive integer x=1

From statement 2
n/x=n
for this to be true x=1

Hence D
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x is a positive integer (given)

From statement 1
x^2=√x --> x^4=x --> x^4-x=0 --> x(x^3-1)=0 -->Since x is a positive integer x^3=1 --> x=1. Hence Sufficient.


From statement 2
n/x=n --> n{(1/x) -1} =0 -->Since n!=0, (1/x)-1=0 -->1/x=1 -->x=1. Hence sufficient.

Hence D
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Hi Bunuel, if we want to follow a factorizarion method I got stucked in the last term when factorizing x (x^3-1) could you advice?
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Bunuel
If x is a positive integer, what is the value of x ?

(1) \(x^2=\sqrt{x}\)

(2) \(\frac{n}{x} = n\) and \(n\neq 0\)


(DS08306)

If x is a positive integer, what is the value of x ?

(1) \(x^2=\sqrt{x}\)
x = 1
SUFFICIENT

(2) \(\frac{n}{x} = n\) and \(n\neq 0\)
x=1
SUFFICIENT

IMO D
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If x is a positive integer, what is the value of x ?

(1) x^2=√x
(2) n/x=n and n≠0

x is a positive number - thus x>0

Statement 1:
x^4 = x
x^4 - x = 0
x(x^3 - 1) = 0
x = 0, 1. but x>0 thus x = 1

Statement 2:
as n is not equal to 0
thus:
1/x=1
x = 1

thus D
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