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Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible
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23 Jan 2015, 18:24
FOIL.
xy + 3x + 2y + 6
2y + 6 must be even (sum of two evens).
xy + 3x must be even (similar logic).
Odd + Odd = Even OR Even + Even = Even
Check Odd+Odd.
Plug in 1 for x. 4xy = 4y --> divisible by 4.
Plug in 2 for x. 4xy = 8y --> divisible by 8. Still divisible by 4.
I am having some trouble with this question. My first thought was that 4 obviously had to be correct because 4xy must be divisible by 4. However, this seems too easy for a 600-700. I guess with different numbers, the answer would be some multiple of 4 rather than 4?
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible
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24 Jan 2015, 00:06
1
ANSWER IS A. Notice WE are told that MUST NOT COULD .
AS we know that (X+2) (Y+3) Is even , X and Y NOT need to be even neccessairly . CONSIDER THIS case : Y and X could be ODD: If X is odd then X+2 is odd too and as
(X+2) (Y+3) is EVEN only one number is to be even and if Y is odd too then Y+3 Is even .so the whole expression is EVEN too.
SO we can have a case in which X and Y could be BOTH odd but the expression is even. in this case we would have 4*X*Y and ONLY is divisible BY 4 .
AS we are told That MUST BE DIVISIBLE ONLY Option A can be fit...
For this Even/Odd number property question, it's important to recognize that if either parenthetical term (x + 2) or (y + 3) is even, then the product of the two will be even, as Even * Odd or Even * Even is even.
This means that the given information could be held true if:
x is even
or
y is odd
If x is even, that means that 4xy would be 4(even)(y), which would allow that product 4xy to be divisible by 8 or more. But if x were odd and y were odd (for example, 1), then 4xy would be 4(odd)(odd), meaning that the only factors of 2 would come from the coefficient of 4. Therefore, as x = 1 and y = 1 are indeed possible, the only answer choice that MUST BE a factor of 4xy is 4.
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible
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04 Feb 2015, 13:55
Hi All,
DesiGmat has caught the 'secret' to this question. Unfortunately, the GMAT isn't likely to make it quite this "easy" on you, but there is something to be said for paying careful attention to what the specific prompt gives you and what the specific question asks for...
For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible
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04 Feb 2015, 18:52
1
DesiGmat wrote:
Bunuel wrote:
For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible by:
A. 4 B. 8 C. 9 D. 10 E. 12
Kudos for a correct solution.
I didn't understand the question.. 4XY will always be divisible by 4 because it already has 4 in it (already multiplied by 4)
why do we have to open any brackets and multiply?
hi desigmat ,
it is not so simple that you dont require to open brackets.... it was simple if no conditions were set as shown in red.. if (x+2)(y+3) is even has to be checked in different scenarios for evenness in this case, you may get the answer without opening the brackets,here are two ways 1) what if it gives you,for eg, x and y are consecutive positive integers .... this would ensure that one of x nd y will be even, so 4xy must be div by 8 in this case.. 2) what if Q says if (x+2)(y+4) is even, and x,y are different integers.. here both x and y have to be even, so 4xy must be div by 8...
What i am trying to tell you is that you have to check a Q for all possibilities before hooking on to an ans
_________________
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74% (01:39) correct 
