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If x, y and z are positive integers such that x^4*y^3 = z^2

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If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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If \(x, y\) and \(z\) are positive integers such that \(x^4y^3 = z^2\), is \(x^9 - y^6\) odd?

(1) \(\frac{x^4y^3}{(x^2 + y^2)}\) can be written in the form \(4k + 3\), where \(k\) is a positive integer

(2) \(z = x + y\)



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Last edited by EgmatQuantExpert on 13 Dec 2016, 05:46, edited 4 times in total.
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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New post 10 Apr 2015, 08:26
Statement 1: 4k+3 is odd, but the nature of fraction can be odd, even or neither depending on the values of x,y & z. Therefore, statement 1 is insufficient.
Statement 2: If we compare the 2 equations: (x^4)(y^3)=z^2 & z=x+y, we see that x,y & z must all be even to satisfy both equations. Hence we know the nature of both x & y, therefore statement 2 is sufficient.
Answer: (B)
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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EgmatQuantExpert wrote:
If \(x, y\) and \(z\) are positive integers such that \(x^4y^3 = z^2\), is \(x^9 - y^6\) odd?

(1) \(\frac{x^4y^3}{(x^2 + y^2)}\) can be written in the form \(4k + 3\), where \(k\) is a positive integer

(2) \(z = x + y\)

We will provide the OA in some time. Till then Happy Solving :lol:

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In this task we know that x, y, z are postive integers so we can eliminate all powers from all equations because they don't have influence on parity
1) we know that \(\frac{xy}{(x+y)}\) equal to odd result (4k + 3). This is possible only if xy and x+y have the same parity.
And they can have the same parity only if they both even.
So if they both even when we can say that z is too even
Sufficient

2) from this statement we know that z = x + y and from task we know that z = xy
This is possible only if xy, (x+y) and z have the same parity.
And they can have the same parity only if they even.
So z is even
Sufficient

Answer is D
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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Detailed Solution

Step-I: Given Info

We are given three positive integers \(x\), \(y\) and \(z\) such that \(x^4y^3 = z^2\), and we are asked to find if \(x^9 – y^6\) is odd

Step-II: Interpreting the Question Statement

For the expression \(x^9 – y^6\) to be odd, both the expressions need to be of opposite even/odd nature i.e. one has to be odd and another has to be even (as even- odd = odd or odd-even = odd). Since the even/odd nature of \(x^9\) would have the same even/odd nature as \(x\) and even/odd nature of \(y^6\) would be the same even/odd nature as \(y\), if we can determine the similar or opposite nature of \(x\) , \(y\) we can determine the even/odd nature of \(x^9 – y^6\).

Step-III: Statement-I

We are told that \(\frac{x^4y^3}{x^2 + y^2}\) can be written in the form \(4k + 3\), where \(k\) is a positive integer. We observe here that a fraction has been simplified to an odd number (as \(4k\) is always even and even + odd = odd) which would imply that both the numerator and the denominator are either even or odd.
Hence, we can say that the product of \(x\) and \(y\) and the sum of \(x\) and \(y\) have the same even/odd nature.

This is possible only if \(x\), \(y\) are both even. Since, we have determined the similar nature of \(x\), \(y\) we can say with certainty that expression \(x^9 – y^6\) is always even.
Hence, statement-I is sufficient to answer the question.

Step-IV: Statement-II

Statement-II tells us that \(z = x + y\), we know that \(z\) is expressed as a product of \(x\) and \(y\). Since the product of \(x\), \(y\) have the same even/odd nature as that of the sum of \(x\) and \(y\), we can say with certainty that \(x\), \(y\) are both even.
Hence, statement-II is sufficient to answer the question.

Step-V: Combining Statements I & II

Since, we have a unique answer from Statement- I & II we don’t need to be combine Statements- I & II.
Hence, the correct answer is Option D

Key Takeaways

1. In even-odd questions, simplify complex expressions into simpler expressions using the properties of even-odd combinations.
2. Know the properties of Even-Odd combinations to save the time spent deriving them in the test
3. The even/odd nature of some expressions can be determined without knowing the exact even/odd nature of the variables of the expressions by using the even/odd combination property


Harley1980- Great Job!
shuvabrata88- You missed out on the analysis of St-I where even/odd nature of a fraction was to be analyzed.

Regards
Harsh
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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We are told that (x^4)*(y^3)/(x^2 + y^2) can be written in the form 4k + 3, where k is a positive integer. We observe here that a fraction has been simplified to an odd number (as 4k is always even and even + odd = odd) which would imply that both the numerator and the denominator are either even or odd.


Hi Harsh,

Could you clarify this. If we take numerator 29 and denominator 4, this can be written as 4*7 + 1 (which is odd). So here, the above rule which you have used doesn't apply? Or am I missing something here?
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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caleb708 wrote:
We are told that (x^4)*(y^3)/(x^2 + y^2) can be written in the form 4k + 3, where k is a positive integer. We observe here that a fraction has been simplified to an odd number (as 4k is always even and even + odd = odd) which would imply that both the numerator and the denominator are either even or odd.


Hi Harsh,

Could you clarify this. If we take numerator 29 and denominator 4, this can be written as 4*7 + 1 (which is odd). So here, the above rule which you have used doesn't apply? Or am I missing something here?


Hi caleb708,

Please note the following points regarding your analysis:

1.\(\frac{29}{4}\) can't be simplified to 4k + 1. Only 29 can be written as 4k + 1. To help you understand better, consider the example \(\frac{33}{3} = 11\) = 4k + 3 where k = 2.

2. A fraction simplifying to an odd expression is only possible if both the numerator and the denominator are either even or odd. Consider the case of opposite even-odd nature of the numerator and the denominator, \(\frac{odd}{even}\) can never be simplified to a integer, for example \(\frac{33}{2}\). Also an \(\frac{even}{odd}\) if simplified to an integer will always yield an even integer, for example \(\frac{36}{3} = 12\).

On the other hand an \(\frac{odd}{odd}\) if simplified to an integer will always yield an odd integer. Similarly an \(\frac{even}{even}\) if simplified to an integer will yield either an even or an odd integer. Hence for a fraction to be simplified to an odd expression, both the numerator and the denominator have to be of the same even-odd nature.

Hope it's clear :)

Regards
Harsh
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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New post 27 May 2015, 06:06
Hi Harsh, can you share an example of when x and y are both even and xy/x+y= odd.. I understand the theory but have not been able to prove it.
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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Naman1987 wrote:
Hi Harsh, can you share an example of when x and y are both even and xy/x+y= odd.. I understand the theory but have not been able to prove it.


Hi Naman1987,

\(\frac{xy}{(x +y)}\) would be odd if \(x+y\) has the same power of 2 as in \(xy\). If the power of 2 in \(x + y\) is less then that in \(xy\), the term \(\frac{xy}{x + y}\) would be even.

Let's assume \(x = y = 6\), then \(xy = 36\) which has \(2^2\) in it. \(x + y = 12\) which also has \(2^2\) in it. So \(\frac{36}{12} = 3\) which is odd.

In statement-I we are talking about various powers of \(x\) and \(y\) in the numerator and the denominator and since power does not have any effect on the even-odd nature of a number we could conclude that product of \(x\) and \(y\) (which is in the numerator) and sum of \(x\) and \(y\)(which is in the denominator) will have the same even-odd nature.

Hope it's clear :)

Regards
Harsh
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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New post 27 May 2015, 23:35
Thanks Harsh! It's very clear now:)
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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New post 19 Jun 2015, 11:25
EgmatQuantExpert wrote:
Detailed Solution

Step-I: Given Info

We are given three positive integers \(x\), \(y\) and \(z\) such that \(x^4y^3 = z^2\), and we are asked to find if \(x^9 – y^6\) is odd

Step-II: Interpreting the Question Statement

For the expression \(x^9 – y^6\) to be odd, both the expressions need to be of opposite even/odd nature i.e. one has to be odd and another has to be even (as even- odd = odd or odd-even = odd). Since the even/odd nature of \(x^9\) would have the same even/odd nature as \(x\) and even/odd nature of \(y^6\) would be the same even/odd nature as \(y\), if we can determine the similar or opposite nature of \(x\) , \(y\) we can determine the even/odd nature of \(x^9 – y^6\).

Step-III: Statement-I

We are told that \(\frac{x^4y^3}{x^2 + y^2}\) can be written in the form \(4k + 3\), where \(k\) is a positive integer. We observe here that a fraction has been simplified to an odd number (as \(4k\) is always even and even + odd = odd) which would imply that both the numerator and the denominator are either even or odd.
Hence, we can say that the product of \(x\) and \(y\) and the sum of \(x\) and \(y\) have the same even/odd nature.

This is possible only if \(x\), \(y\) are both even. Since, we have determined the similar nature of \(x\), \(y\) we can say with certainty that expression \(x^9 – y^6\) is always even.
Hence, statement-I is sufficient to answer the question.

Step-IV: Statement-II

Statement-II tells us that \(z = x + y\), we know that \(z\) is expressed as a product of \(x\) and \(y\). Since the product of \(x\), \(y\) have the same even/odd nature as that of the sum of \(x\) and \(y\), we can say with certainty that \(x\), \(y\) are both even.
Hence, statement-II is sufficient to answer the question.


Step-V: Combining Statements I & II

Since, we have a unique answer from Statement- I & II we don’t need to be combine Statements- I & II.
Hence, the correct answer is Option D

Key Takeaways

1. In even-odd questions, simplify complex expressions into simpler expressions using the properties of even-odd combinations.
2. Know the properties of Even-Odd combinations to save the time spent deriving them in the test
3. The even/odd nature of some expressions can be determined without knowing the exact even/odd nature of the variables of the expressions by using the even/odd combination property


Harley1980- Great Job!
shuvabrata88- You missed out on the analysis of St-I where even/odd nature of a fraction was to be analyzed.

Regards
Harsh


hi Harsh..! how did you work out the 2nd statement..? after reading ur solution and thinking on it, now it makes sense. But initially it was just difficlut to strike.
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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riyazgilani wrote:
EgmatQuantExpert wrote:
Detailed Solution

Step-I: Given Info

We are given three positive integers \(x\), \(y\) and \(z\) such that \(x^4y^3 = z^2\), and we are asked to find if \(x^9 – y^6\) is odd

Step-II: Interpreting the Question Statement

For the expression \(x^9 – y^6\) to be odd, both the expressions need to be of opposite even/odd nature i.e. one has to be odd and another has to be even (as even- odd = odd or odd-even = odd). Since the even/odd nature of \(x^9\) would have the same even/odd nature as \(x\) and even/odd nature of \(y^6\) would be the same even/odd nature as \(y\), if we can determine the similar or opposite nature of \(x\) , \(y\) we can determine the even/odd nature of \(x^9 – y^6\).

Step-III: Statement-I

We are told that \(\frac{x^4y^3}{x^2 + y^2}\) can be written in the form \(4k + 3\), where \(k\) is a positive integer. We observe here that a fraction has been simplified to an odd number (as \(4k\) is always even and even + odd = odd) which would imply that both the numerator and the denominator are either even or odd.
Hence, we can say that the product of \(x\) and \(y\) and the sum of \(x\) and \(y\) have the same even/odd nature.

This is possible only if \(x\), \(y\) are both even. Since, we have determined the similar nature of \(x\), \(y\) we can say with certainty that expression \(x^9 – y^6\) is always even.
Hence, statement-I is sufficient to answer the question.

Step-IV: Statement-II

Statement-II tells us that \(z = x + y\), we know that \(z\) is expressed as a product of \(x\) and \(y\). Since the product of \(x\), \(y\) have the same even/odd nature as that of the sum of \(x\) and \(y\), we can say with certainty that \(x\), \(y\) are both even.
Hence, statement-II is sufficient to answer the question.


Step-V: Combining Statements I & II

Since, we have a unique answer from Statement- I & II we don’t need to be combine Statements- I & II.
Hence, the correct answer is Option D

Key Takeaways

1. In even-odd questions, simplify complex expressions into simpler expressions using the properties of even-odd combinations.
2. Know the properties of Even-Odd combinations to save the time spent deriving them in the test
3. The even/odd nature of some expressions can be determined without knowing the exact even/odd nature of the variables of the expressions by using the even/odd combination property


Harley1980- Great Job!
shuvabrata88- You missed out on the analysis of St-I where even/odd nature of a fraction was to be analyzed.

Regards
Harsh


hi Harsh..! how did you work out the 2nd statement..? after reading ur solution and thinking on it, now it makes sense. But initially it was just difficlut to strike.


Hi riyazgilani,

Statement-II plays on the properties of sum and product of two even/odd numbers. Refer the below diagram wherein we assume all possible combinations of two numbers being even/odd and analyze the even/odd nature of their sum and product.


Image


You would observe that the product and sum of two numbers have the same even/odd nature only if both are even. In the question, z is expressed as sum of x and y as well as the product of x and y. Hence, x + y and xy should have the same even/odd nature. This is only possible if x and y both are even.

Hope it's clear :)

Regards
Harsh
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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New post 29 Jul 2015, 12:42
Harsh,

What I have understood so far after analyzing the question stem and taking out the inference is this I need to prove
x-y =odd ; for that both of them have to be different parity one needs to be odd and other needs to be even.

In statement 1 : xy/x+y =odd , xy can be odd for eg. 9 and x+y can also be odd i.e. 3 and result is 3 which is odd and satisfying the equation, why we have marked both of them as even ( in your solution ).
If some how they ( xy and x+y ) are even how can x and y individually are even.
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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New post 29 Jul 2015, 13:17
shekharshrek wrote:
Harsh,

What I have understood so far after analyzing the question stem and taking out the inference is this I need to prove
x-y =odd ; for that both of them have to be different parity one needs to be odd and other needs to be even.

In statement 1 : xy/x+y =odd , xy can be odd for eg. 9 and x+y can also be odd i.e. 3 and result is 3 which is odd and satisfying the equation, why we have marked both of them as even ( in your solution ).
If some how they ( xy and x+y ) are even how can x and y individually are even.


You are correct about the text in green.

Look below for statements in red:

xy/x+y = odd --- it can either be of the form 6/2 or 9/3 . You have already covered the case for 9/3 giving x and y of same nature (odd-odd or even-even).

When x+y =even ---> x and y can either be (even, even) or (odd, odd) but as we have been given that xy = even ---> x,y MUST be (even,even) [Odd * Odd \(\neq\) even number , think 3*9 = 27 or 3*7 = 21 etc.] and the reason that these statements are still sufficient is even - even = even and odd - odd = even.

Hope this helps.
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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shekharshrek wrote:
Harsh,

What I have understood so far after analyzing the question stem and taking out the inference is this I need to prove
x-y =odd ; for that both of them have to be different parity one needs to be odd and other needs to be even.

In statement 1 : xy/x+y =odd , xy can be odd for eg. 9 and x+y can also be odd i.e. 3 and result is 3 which is odd and satisfying the equation, why we have marked both of them as even ( in your solution ).
If some how they ( xy and x+y ) are even how can x and y individually are even.


Hi shekharshrek,

If the product of two numbers x and y is even that would mean that at least one of them is even. Hence the following combinations may occur:

I. x is even and y is odd- In this case xy is even and x + y is odd

II. x is odd and y is even - In this case xy is even and x + y is odd

III. x and y both are even - In this case xy is even and x + y is also even.

So, if the product and the sum of two number have the same even/odd nature, both the numbers would be even.You can also refer this post, which explains the same point.

Hope this helps :)

Regards
Harsh
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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[quote="EgmatQuantExpert"]Detailed Solution


Step-III: Statement-I

We are told that \(\frac{x^4y^3}{x^2 + y^2}\) can be written in the form \(4k + 3\), where \(k\) is a positive integer. We observe here that a fraction has been simplified to an odd number (as \(4k\) is always even and even + odd = odd) which would imply that both the numerator and the denominator are either even or odd.
Hence, we can say that the product of \(x\) and \(y\) and the sum of \(x\) and \(y\) have the same even/odd nature.

This is possible only if \(x\), \(y\) are both even. Since, we have determined the similar nature of \(x\), \(y\) we can say with certainty that expression \(x^9 – y^6\) is always even.
Hence, statement-I is sufficient to answer the question.

Hi Harsh,

Can you please elaborate on statement 1? I understand that the result of the fraction is an odd number so both the denominator and the numerator can be either even or odd. But how did you come to the statement that "the product of \(x\) and \(y\) and the sum of \(x\) and \(y\) have the same even/odd nature"?
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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New post 26 Aug 2015, 03:28
harishbiyani8888 wrote:
EgmatQuantExpert wrote:
Detailed Solution


Step-III: Statement-I

We are told that \(\frac{x^4y^3}{x^2 + y^2}\) can be written in the form \(4k + 3\), where \(k\) is a positive integer. We observe here that a fraction has been simplified to an odd number (as \(4k\) is always even and even + odd = odd) which would imply that both the numerator and the denominator are either even or odd.
Hence, we can say that the product of \(x\) and \(y\) and the sum of \(x\) and \(y\) have the same even/odd nature.

This is possible only if \(x\), \(y\) are both even. Since, we have determined the similar nature of \(x\), \(y\) we can say with certainty that expression \(x^9 – y^6\) is always even.
Hence, statement-I is sufficient to answer the question.

Hi Harsh,

Can you please elaborate on statement 1? I understand that the result of the fraction is an odd number so both the denominator and the numerator can be either even or odd. But how did you come to the statement that "the product of \(x\) and \(y\) and the sum of \(x\) and \(y\) have the same even/odd nature"?


Let me answer your question. Refer to if-x-y-and-z-are-positive-integers-such-that-x-4-y-3-z-196033.html#p1554972 in which I show that when you have a scenario:

a/b = odd then, the only possible cases are : 6/2 or 9/3 in which case both 6 or 2 are even and 9 or 3 are odd. So you see the numerator and the denominator have the same 'nature'.

Hope this helps.
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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New post 05 Jan 2016, 22:23
EgmatQuantExpert (and others who can help)

I've done all 10 questions, scoring roughly 60% correct including a couple "lucky guesses". What would be your next suggestions for practice? The Quant OG Guide problems seem too simple because I rarely learn anything from study sessions with the materials, learning a method or two per 25/30 questions. This is the same problem I run into with many other sources I've used and I believe this cannot be the most efficient method to improve. Do you have any suggestions? (other than of course buying more and more materials)
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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New post 06 Jan 2016, 17:45
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x,y and z are positive integers such that x 4 y 3 =z 2 , is x 9 −y 6 odd?

(1) x 4 y 3 (x 2 +y 2 ) can be written in the form 4k+3 , where k is a positive integer

(2) z=x+y


In the original condition, there are 3 variables(x,y,z) and 1 equations(x^4y^3=z^2), which should match with the number of equations. So you need 2 more equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1)&2), in 2), from z=x+y, x=odd, y=odd, z=even(not possible) and x=odd, y=even z=odd(also not possible). From x=even, y=even, z=even -> x^9-y^6=even-even=even, which is no and sufficient.
In 1), x^4y^3/(x^2+y^2)=4k+3=odd also becomes x^4y^3=odd(x^2+y^2). What satisfies this is x=z=even, which is no and sufficient. Therefore, the answer is D. This is the mistake type 4.


 For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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New post 06 Jan 2016, 20:58
Thanks MathRevolution. After reviewing these questions a few days later (and after rest and recovery), the solutions and questions make a whole lot more sense, except for maybe 8 (this question seems absurd. Maybe because the answer is e and on the actual exam there's no need to find every case to realize both statements aren't sufficient even together).
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2 [#permalink]

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New post 23 Jan 2018, 06:31
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Re: If x, y and z are positive integers such that x^4*y^3 = z^2   [#permalink] 23 Jan 2018, 06:31
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