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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If x and y are distinct positive integers, is |x − y| a factor of 12?

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Math Expert V
Joined: 02 Sep 2009
Posts: 59590
If x and y are distinct positive integers, is |x − y| a factor of 12?  [#permalink]

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1
6 00:00

Difficulty:   95% (hard)

Question Stats: 32% (02:28) correct 68% (02:17) wrong based on 124 sessions

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If x and y are distinct positive integers, is |x − y| a factor of 12?

(1) $$x^2 − 6x + y^2 − 4y = 0$$

(2) x = 1

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Math Expert V
Joined: 02 Aug 2009
Posts: 8290
If x and y are distinct positive integers, is |x − y| a factor of 12?  [#permalink]

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3
1
Bunuel wrote:
If x and y are distinct positive integers, is |x − y| a factor of 12?

(1) $$x^2 − 6x + y^2 − 4y = 0$$

(2) x = 1

Great question Bunuel..
Noone has answered in last 3 hours shows it is tricky..

(1) $$x^2 − 6x + y^2 − 4y = 0$$
$$x^2 − 6x + y^2 − 4y = 0........x^2-6x+9+y^2-4y+4=9+4......(x-3)^2+(y-2)^2=13=4+9$$
Only possible values is 4+9 as we have to find 13 as sum of two squares - 4+9
Two ways
a) $$x-3=x-3=√9=3......x=6$$ AND $$y-2=√4=2$$.....$$y=4$$ therefore $$|x-y|=6-4=2$$..yes a factor of 12
b) $$x-3=√4.....x=5$$ AND $$y-2=√9=3...y=5$$ ...BUT x and y are distinct.. so not possible
Ans yes
Sufficient

(2) x = 1
Insufficient

A
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Re: If x and y are distinct positive integers, is |x − y| a factor of 12?  [#permalink]

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E

1 ---->
x(x-6) + y(y-6) = 0
four cases

x=0 y= 0 No
x=6 y =0 Yes
x=0 y=4 Yes
x= 6 y = 4 Yes
Insuff.

2. x=1 Insuffcient
1+2----. x=1 that means y = -1 or 5
----> mod(x-y) = 0 or 4
Insufficient.

Senior Manager  G
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Re: If x and y are distinct positive integers, is |x − y| a factor of 12?  [#permalink]

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chetan2u wrote:
Bunuel wrote:
If x and y are distinct positive integers, is |x − y| a factor of 12?

(1) $$x^2 − 6x + y^2 − 4y = 0$$

(2) x = 1

Great question Bunuel..
Noone has answered in last 3 hours shows it is tricky..

(1) $$x^2 − 6x + y^2 − 4y = 0$$
$$x^2 − 6x + y^2 − 4y = 0........x^2-6x+9+y^2-4y+4=9+4......(x-3)^2+(y-2)^2=13=4+9$$
Only possible values is 4+9 as we have to find 13 as sum of two squares - 4+9
Two ways
a) x-3=x-3=√9=3......x=6 AND y-2=√4=2.....y=4 therefore|x-y|=6-4=2..yes a factor of 12
b) x-3=√4.....x=5 AND y-2=√9=3...y=5 ...BUT x and y are distinct.. so not possible
Ans yes
Sufficient

(2) x = 1
Insufficient

A

Hi !

How did you come to sum of squares will equal 13 ?
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Re: If x and y are distinct positive integers, is |x − y| a factor of 12?  [#permalink]

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This is definitely a question where the math is pretty interesting. If you are a math person, the danger here, strategically, is getting sucked into the fun of the math and wasting three minutes. If you are not a math person, the danger here is staring at statement one for three minutes and end up guessing in a panic.

So, this is how I recommend approaching a question like this. First, evaluate the easier statement first. Statement two is NOT sufficient, so eliminate B and D.

Second, set yourself a time budget to evaluate the harder statement. Whatever strategy you choose - math or example numbers being (probably) the two primary strategies - can take a lot of time and you don't want to spend too much time choosing between A, C, and E.

For statement one this is what I would do. First, I would rearrange the equation to look like this:

x(x-6) + y(y-4) = 0

There are two ways that this statement could equal zero.

1. Both parts equal 0 meaning x(x-6)=0 and y(y-4)=0
2. One is positive and the other is negative but with the same absolute value.

Take the easier of the two situations first. If this is the case, x = 6 and y = 4, so |x - y| = |6-4| = 2, which is a factor of 12. So, statement one works at least sometimes.

By this time, you've likely spent at least 2 minutes. Working through 2 (one positive and one negative) is going to take a lot of math or example numbers. DON'T DO THIS. At this point, guess among A, C, and E and move on.

If you are really ahead on time, the arithmetic is probably easier than the math. Make a table and pick some numbers to see what works. I came up with

x y
1 5
6 4 (from above)

This list may not be exhaustive, but for the purposes of the GMAT it is probably fine. For these, |x-y| is a factor of 12. Go with A and move on.
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Jayson Beatty
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Math Expert V
Joined: 02 Aug 2009
Posts: 8290
Re: If x and y are distinct positive integers, is |x − y| a factor of 12?  [#permalink]

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ShankSouljaBoi wrote:
E

1 ---->
x(x-6) + y(y-6) = 0
four cases

x=0 y= 0 No
x=6 y =0 Yes
x=0 y=4 Yes
x= 6 y = 4 Yes
Insuff.

2. x=1 Insuffcient
1+2----. x=1 that means y = -1 or 5
----> mod(x-y) = 0 or 4
Insufficient.

Firstly the MISTAKE ...
x and y are distinct + positive +integer
in three cases x or y or both are 0 but 0 is not positive so discard those pairs..
ONLY 6,4 are left....
hence A is suff

NOW on your question how square are 13
$$x^2 − 6x + y^2 − 4y = 0........x^2-6x+9+y^2-4y+4=9+4......(x-3)^2+(y-2)^2=13=4+9$$
x^2-6x is of the form a^2-2ab+b^2=(a-b)^2 so a is x, b becomes 3....$$x^2-2*3*x+3^2=(x-3)^2$$ so we add 3^2 on BOTH sides and similarly 2^2 with y^2-4y
combined 9+4 on both sides
Only possible values is 4+9 as we have to find 13 as sum of two squares - 4+9
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Re: If x and y are distinct positive integers, is |x − y| a factor of 12?  [#permalink]

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Thank you !! What a blunder Manager  G
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Re: If x and y are distinct positive integers, is |x − y| a factor of 12?  [#permalink]

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Bunuel wrote:
If x and y are distinct positive integers, is |x − y| a factor of 12?

(1) $$x^2 − 6x + y^2 − 4y = 0$$

(2) x = 1

Experts,I want some suggestion from you on this approach:
Since statement 2 seem to be simple i used that and substituted it in statement 1.Statement 1 involved 2 variables and 1 equation and so would have many possible roots.I did not add 9 and 4 as chetan2u has done as that did not occur to me.By solving for x=1 ,i get y=5 or y=1.Since distinct and positive y=5. Re: If x and y are distinct positive integers, is |x − y| a factor of 12?   [#permalink] 07 Oct 2018, 21:29
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