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What is the absolute difference between the cubes of two different non-negative integers?

Given: \(x\) and \(y\) are different non-negative integers. Question: \(|x^3-y^3|=?\)

(1) One of the integers is 2 greater than the other integer --> \(|x-y|=2\). There exist infinitely many such \(x\) and \(y\), thus we'll have infinitely many different values of \(|x^3-y^3|\). For example consider \(x=2\), \(y=0\) and \(x=3\), \(y=1\). Not sufficient.

(2) The square of the sum of the integers is 49 greater than the product of the integers --> \((x+y)^2=xy+49\) --> \(x^2+xy+y^2=49\). Now, we can not simply say that this statement is not sufficient (as done in the above post), because we have some constraints on \(x\) and \(y\), namely that they are non-negative integers. Taking this into account, from \(x^2+xy+y^2=49\) we can say that both \(x\) and \(y\) are integers from 0 to 7, inclusive. By trial and error we can easily find that only two pairs satisfy this equation: (7,0) and (5,3), in any order. Not sufficient.

Notice that if we were told that \(x\) and \(y\) are positive integers (instead of non-negative integers) then this statement would be sufficient, since only one pair would remain: (5,3).

(1)+(2) Since from (1) \(|x-y|=2\) then from (2) only one pair is valid: (5,3). Sufficient.

Re: What is the absolute difference between the cubes of two [#permalink]

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16 Mar 2012, 04:04

IMO the correct answer is C first the requirement is to find the value of |x^3 - y^3| or |(x-y)( x^2 + y^2 + xy)| 1) x = y + 2 or x -y = 2, Not sufficient 2) (x+y)^2 = 49 + xy, as we simplify we get x^2 + y^2 + xy = 49 insufficient because we don't know the value of x - y from this statement combine 1 and 2, we get the value of (x-y) and ( x^2 + y^2 + xy) hence sufficient.
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Re: What is the absolute difference between the cubes of two [#permalink]

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16 Mar 2012, 04:08

Hey Bunnel, thanx for a nice explanation . I have learned something new today with ur explanation. Every time i read ur posts, it further reimposes my faith that ur a true legend. Thanx a ton.
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Fire the final bullet only when you are constantly hitting the Bull's eye, till then KEEP PRACTICING.

A WAY TO INCREASE FROM QUANT 35-40 TO 47 : http://gmatclub.com/forum/a-way-to-increase-from-q35-40-to-q-138750.html

Q 47/48 To Q 50 + http://gmatclub.com/forum/the-final-climb-quest-for-q-50-from-q47-129441.html#p1064367

Three good RC strategies http://gmatclub.com/forum/three-different-strategies-for-attacking-rc-127287.html

Could you please help me to understand where is the flaw in my reasoning: I chose the answer D for this problem, since the first statement gives us the relationship between two variables : if the second item is two more than the first one, doesn't it mean x=x+2, where x+2 =y

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Could you please help me to understand where is the flaw in my reasoning: I chose the answer D for this problem, since the first statement gives us the relationship between two variables : if the second item is two more than the first one, doesn't it mean x=x+2, where x+2 =y

Merging similar topics. Please refer to the solutions above and ask if anything remains unclear.
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Re: What is the absolute difference between the cubes of two [#permalink]

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04 Apr 2012, 11:21

But, Bunuel, why can't i set x=x+2, where x+2=y? Since the first statement tells us "one of the integers is 2 greater than the other". The only thing confused me - "ONE of the integers", what means i don't know the exact order May be my question is stupid one- but i just need to grasp it fully

But, Bunuel, why can't i set x=x+2, where x+2=y? Since the first statement tells us "one of the integers is 2 greater than the other". The only thing confused me - "ONE of the integers", what means i don't know the exact order May be my question is stupid one- but i just need to grasp it fully

First of all x=x+2 doesn't make any sense, because x's cancel out and we have that 0=2, which is not true.

Next, "one of the integers is 2 greater than the other integer" CAN be expressed as x+2=y, (assuming that y is larger), but even in this case we cannot get the single numerical value of y^3-x^3, for example consider: y=2, x=0 and y=3, x=1.

But are we expected to know 5,3 is the pair which solves given condition if integers are only positive?

You can get that the statements taken together are sufficient even if you don't know that the only pair possible is (5, 3): \(x^3-y^3=(x-y) (x^2+x y+y^2)\) --> (1) says that \(x-y=2\) and (2) says that \(x^2+x y+y^2=49\) --> \(x^3-y^3=(x-y) (x^2+x y+y^2)=2*49=98\).

Re: What is the absolute difference between the cubes of two [#permalink]

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29 Nov 2013, 07:29

Hello from the GMAT Club BumpBot!

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Re: What is the absolute difference between the cubes of two [#permalink]

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09 Jul 2014, 01:57

Bunuel wrote:

vdbhamare wrote:

IMO C..

nice work bunnel.....

But are we expected to know 5,3 is the pair which solves given condition if integers are only positive?

You can get that the statements taken together are sufficient even if you don't know that the only pair possible is (5, 3): \(x^3-y^3=(x-y) (x^2+x y+y^2)\) --> (1) says that \(x-y=2\) and (2) says that \(x^2+x y+y^2=49\) --> \(x^3-y^3=(x-y) (x^2+x y+y^2)=2*49=98\).

Hope it's clear.

It was difficult to see that \(x^3-y^3\) can we written as \((x-y) (x^2+x y+y^2)\)

is there an easier way to understand this or must it be learnt by heart.

But are we expected to know 5,3 is the pair which solves given condition if integers are only positive?

You can get that the statements taken together are sufficient even if you don't know that the only pair possible is (5, 3): \(x^3-y^3=(x-y) (x^2+x y+y^2)\) --> (1) says that \(x-y=2\) and (2) says that \(x^2+x y+y^2=49\) --> \(x^3-y^3=(x-y) (x^2+x y+y^2)=2*49=98\).

Hope it's clear.

It was difficult to see that \(x^3-y^3\) can we written as \((x-y) (x^2+x y+y^2)\)

is there an easier way to understand this or must it be learnt by heart.

Re: What is the absolute difference between the cubes of two [#permalink]

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11 Jul 2014, 05:05

Bunuel wrote:

What is the absolute difference between the cubes of two different non-negative integers?

Given: \(x\) and \(y\) are different non-negative integers. Question: \(|x^3-y^3|=?\)

(1) One of the integers is 2 greater than the other integer --> \(|x-y|=2\). There exist infinitely many such \(x\) and \(y\), thus we'll have infinitely many different values of \(|x^3-y^3|\). For example consider \(x=2\), \(y=0\) and \(x=3\), \(y=1\). Not sufficient.

(2) The square of the sum of the integers is 49 greater than the product of the integers --> \((x+y)^2=xy+49\) --> \(x^2+xy+y^2=49\). Now, we can not simply say that this statement is not sufficient (as done in the above post), because we have some constraints on \(x\) and \(y\), namely that they are non-negative integers. Taking this into account, from \(x^2+xy+y^2=49\) we can say that both \(x\) and \(y\) are integers from 0 to 7, inclusive. By trial and error we can easily find that only two pairs satisfy this equation: (7,0) and (5,3), in any order. Not sufficient.

Notice that if we were told that \(x\) and \(y\) are positive integers (instead of non-negative integers) then this statement would be sufficient, since only one pair would remain: (5,3).

(1)+(2) Since from (1) \(|x-y|=2\) then from (2) only one pair is valid: (5,3). Sufficient.

Answer: C.

Hope it's clear.

Hello Bunuel . After solving stem 2 and coming to the point of \(x^2+xy+y^2=49\) , I deduced that since it is an equation in 2 variables it will not be solvable. I substituted from 1 ( y=x+2) and then realized that it transforms to a simple quadratic and then with the mentioned restriction , came to C. Would this be a right approach ?

What is the absolute difference between the cubes of two different non-negative integers?

Given: \(x\) and \(y\) are different non-negative integers. Question: \(|x^3-y^3|=?\)

(1) One of the integers is 2 greater than the other integer --> \(|x-y|=2\). There exist infinitely many such \(x\) and \(y\), thus we'll have infinitely many different values of \(|x^3-y^3|\). For example consider \(x=2\), \(y=0\) and \(x=3\), \(y=1\). Not sufficient.

(2) The square of the sum of the integers is 49 greater than the product of the integers --> \((x+y)^2=xy+49\) --> \(x^2+xy+y^2=49\). Now, we can not simply say that this statement is not sufficient (as done in the above post), because we have some constraints on \(x\) and \(y\), namely that they are non-negative integers. Taking this into account, from \(x^2+xy+y^2=49\) we can say that both \(x\) and \(y\) are integers from 0 to 7, inclusive. By trial and error we can easily find that only two pairs satisfy this equation: (7,0) and (5,3), in any order. Not sufficient.

Notice that if we were told that \(x\) and \(y\) are positive integers (instead of non-negative integers) then this statement would be sufficient, since only one pair would remain: (5,3).

(1)+(2) Since from (1) \(|x-y|=2\) then from (2) only one pair is valid: (5,3). Sufficient.

Answer: C.

Hope it's clear.

Hello Bunuel . After solving stem 2 and coming to the point of \(x^2+xy+y^2=49\) , I deduced that since it is an equation in 2 variables it will not be solvable. I substituted from 1 ( y=x+2) and then realized that it transforms to a simple quadratic and then with the mentioned restriction , came to C. Would this be a right approach ?

Have you read this: Now, we can not simply say that this statement is not sufficient (as done in the above post), because we have some constraints on \(x\) and \(y\), namely that they are non-negative integers. Taking this into account, from \(x^2+xy+y^2=49\) we can say that both \(x\) and \(y\) are integers from 0 to 7, inclusive. By trial and error we can easily find that only two pairs satisfy this equation: (7,0) and (5,3), in any order. Not sufficient.

Notice that if we were told that \(x\) and \(y\) are positive integers (instead of non-negative integers) then this statement would be sufficient, since only one pair would remain: (5,3). _________________

Re: What is the absolute difference between the cubes of two [#permalink]

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16 Sep 2015, 03:32

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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