Last visit was: 26 Apr 2024, 03:05 It is currently 26 Apr 2024, 03:05

Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Kudos
Math Expert
Joined: 02 Sep 2009
Posts: 92929
Own Kudos [?]: 619101 [14]
Given Kudos: 81609
Send PM
avatar
Manager
Manager
Joined: 27 Oct 2013
Posts: 176
Own Kudos [?]: 225 [2]
Given Kudos: 79
Location: India
Concentration: General Management, Technology
GMAT Date: 03-02-2015
GPA: 3.88
Send PM
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11179
Own Kudos [?]: 31937 [1]
Given Kudos: 290
Send PM
Senior Manager
Senior Manager
Joined: 28 Feb 2014
Posts: 269
Own Kudos [?]: 318 [1]
Given Kudos: 132
Location: United States
Concentration: Strategy, General Management
Send PM
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible [#permalink]
1
Kudos
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.


Just looking at the question, (x+2)(y+3) could be even*even, odd*even, or even*odd in order for the whole expression to be even.

However, just looking at 4xy, since x*y will always be an integer, it will always be divisible by 4. 4 divided by 4 is 1, hence preserving x*y.

Answer: A. 4
avatar
Manager
Manager
Joined: 13 Dec 2013
Posts: 51
Own Kudos [?]: 82 [1]
Given Kudos: 35
Location: Iran (Islamic Republic of)
Send PM
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible [#permalink]
1
Kudos
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...
avatar
Intern
Intern
Joined: 09 Apr 2013
Posts: 27
Own Kudos [?]: 30 [1]
Given Kudos: 15
Schools: Booth '16
Send PM
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible [#permalink]
1
Kudos
here is my explanation:
Given that (x+2)*(y+3) is even. therefore here are the possible cases:

(x+2) (y+3) (x+2) (y+3) (x+2) (y+3)
E E E O O E

x y x y x y
E O E E O O

xy xy xy
E E O

can be divided can be divided must only be
by 4, 8 etc... by 4, 8 etc... divided by 4

Hence A
User avatar
Manager
Manager
Joined: 03 Oct 2014
Posts: 115
Own Kudos [?]: 96 [1]
Given Kudos: 89
Location: India
Concentration: Operations, Technology
GMAT 1: 720 Q48 V40
WE:Engineering (Aerospace and Defense)
Send PM
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible [#permalink]
1
Kudos
Solving Algebraically.....

xy + 3x + 2y + 6 = 2k
xy = 2k - 3x - 2y - 6
4xy = 8k - 12x - 8y - 24....

Now with least value of x,y as 1,1 the condition of being even will suffice..

So only surity is 4xy divisible by 4.
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11179
Own Kudos [?]: 31937 [1]
Given Kudos: 290
Send PM
For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible [#permalink]
1
Kudos
Expert Reply
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
avatar
Manager
Manager
Joined: 01 May 2013
Posts: 57
Own Kudos [?]: 84 [0]
Given Kudos: 8
Send PM
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible [#permalink]
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?
Math Expert
Joined: 02 Sep 2009
Posts: 92929
Own Kudos [?]: 619101 [0]
Given Kudos: 81609
Send PM
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible [#permalink]
Expert Reply
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.



VERITAS PREP OFFICIAL SOLUTION:

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.

Answer: A.
avatar
SVP
SVP
Joined: 27 Dec 2012
Status:The Best Or Nothing
Posts: 1562
Own Kudos [?]: 7208 [0]
Given Kudos: 193
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Send PM
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible [#permalink]
Answer = A= 4

Let x = 3 & y = 1

(x+2)(y+3) would be even with the above values

4xy = 12 would be divisible by 4
GMAT Club Legend
GMAT Club Legend
Joined: 19 Dec 2014
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Posts: 21846
Own Kudos [?]: 11668 [0]
Given Kudos: 450
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Send PM
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible [#permalink]
Expert Reply
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...

GMAT assassins aren't born, they're made,
Rich
Alum
Joined: 12 Aug 2015
Posts: 2282
Own Kudos [?]: 3131 [0]
Given Kudos: 893
GRE 1: Q169 V154
Send PM
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible [#permalink]
No need to solve this one
4xy must be divisible by 4xy for x and y are integers
Smash that A
Current Student
Joined: 10 Sep 2019
Posts: 137
Own Kudos [?]: 33 [0]
Given Kudos: 59
Location: India
Concentration: Social Entrepreneurship, Healthcare
GMAT 1: 680 Q49 V33
GMAT 2: 720 Q50 V37
GRE 1: Q167 V159
GPA: 2.59
WE:Project Management (Non-Profit and Government)
Send PM
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible [#permalink]
Irrespective of the integer value of x & y, '4' * X*Y will always be divisible by 4.

Ans A
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3726
Own Kudos [?]: 16843 [0]
Given Kudos: 165
Send PM
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible [#permalink]
Expert Reply

Solution



Given
    • For integers x and y, (x+2)(y+3) is even.

To find
    • 4xy must be divisible by which number?

Approach and Working out

    • (x+2)(y+3) = Even
      o Case 1 - ) x+2 = Even and y +3 = Odd
         x = Even and y= Odd
          • 4xy will be definitely divisible by 8

      o Case 2 - ) x+2 = Odd and y +3 = Even
         x = Odd and y= Odd
          • 4xy will be definitely divisible by 4

In both the cases, 4xy is definitely divisible by 4.

Thus, option A is the correct answer.
Correct Answer: Option A
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 32678
Own Kudos [?]: 822 [0]
Given Kudos: 0
Send PM
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible [#permalink]
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.
GMAT Club Bot
Re: For integers x and y, if (x+2)(y+3) is even then 4xy must be divisible [#permalink]
Moderators:
Math Expert
92929 posts
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne