BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Manhattan Question_Puzzled

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
User avatar
Community Manager
Posts: 1060
Joined: Fri May 13, 2011 6:46 am
Location: Utrecht, The Netherlands
Thanked: 318 times
Followed by:52 members

by neelgandham » Tue Nov 15, 2011 10:32 pm
If X, Y and z are integers and xyz is divisible by 8, is x even?
(1) yz is divisible by 4
If the value of yz = 4, then x should be even
If the value of yz = 8, then x can be odd or even
Doesn't answer the question, Insufficient!
(2) X, Y, and z are all NOT divisible by 4
If x , y , z are multiples of 2, then xyz is divisible by 8 and x is even. This is the only condition which satisfies the conditions xyz divisible by 8 and x,y,z are all not divisible by 4
Answers the question, Sufficient!

Wooo.. tricky !

IMO B
Last edited by neelgandham on Wed Nov 16, 2011 3:06 am, edited 3 times in total.
Anil Gandham
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/
Top
Master | Next Rank: 500 Posts
Posts: 416
Joined: Thu Jul 28, 2011 12:48 am
Thanked: 28 times
Followed by:6 members

by gunjan1208 » Tue Nov 15, 2011 10:35 pm
Hi Neel,

What about 136? It is divisible by 8 and suffice for condition B also.....But the X is odd
Top
User avatar
Community Manager
Posts: 1060
Joined: Fri May 13, 2011 6:46 am
Location: Utrecht, The Netherlands
Thanked: 318 times
Followed by:52 members

by neelgandham » Tue Nov 15, 2011 10:38 pm
gunjan1208 wrote:Hi Neel,

What about 136? It is divisible by 8 and suffice for condition B also.....But the X is odd
136 ? Are you talking about the value of xyz = 136 ? even if it is 136, the question asked is 'Is X even?'. You can split 136 in as many ways as you can (satisfying the 2nd condition), you will land with the x being an even number

eg x y z are all NOT divisible by 4

x = 2, y = 2, z = 34
x = 34 y = 2, z = 2
x = 2, y = 34, z = 2

are the only three cases! Hope it helps :)
Anil Gandham
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/
Top
Master | Next Rank: 500 Posts
Posts: 416
Joined: Thu Jul 28, 2011 12:48 am
Thanked: 28 times
Followed by:6 members

by gunjan1208 » Tue Nov 15, 2011 11:17 pm
No Neel,

I am not getting it.....This question has eaten 2 hours already from me....

XYZ =136, I presume that X=1, Is this right? Or do I need to factor them?

Can you explain it in better language?
Top
User avatar
Community Manager
Posts: 1060
Joined: Fri May 13, 2011 6:46 am
Location: Utrecht, The Netherlands
Thanked: 318 times
Followed by:52 members

by neelgandham » Tue Nov 15, 2011 11:25 pm
gunjan1208 wrote:No Neel,

I am not getting it.....This question has eaten 2 hours already from me....

XYZ =136, I presume that X=1, Is this right? Or do I need to factor them?

Can you explain it in better language?
if x = 1 then the value of y, z can be ?

y = 134 and z = 1
y = 68 and z = 2
y = 34 and z = 4
y = 17 and z = 8
y = 1 and z = 134
y = 2 and z = 68
y = 4 and z = 34
y = 8 and z = 17

None of the cases satisfy the condition
x,y,and z are all NOT divisible by 4,
Why ? see below

y = 134 and z = 1,Here y is divisible by 4
y = 68 and z = 2, Here y is divisible by 4
y = 34 and z = 4, Here Z is divisible by 4
y = 17 and z = 8, Here Z is divisible by 4
y = 1 and z = 134, Here Z is divisible by 4
y = 2 and z = 68, Here Z is divisible by 4
y = 4 and z = 34, Here y is divisible by 4
y = 8 and z = 17, Here y is divisible by 4

So your assumption is wrong as it doesn't satisfy the condition. You got to assume the values of x,y,z in such a way that it satisfies the condition specified.
Anil Gandham
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/
Top
Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 am
Thanked: 230 times
Followed by:21 members

by shankar.ashwin » Wed Nov 16, 2011 12:11 am
Another way of looking at it.

8 has 3 2's (2*2*2) - We want to find if one of these 2's is in X ( Is X even?)

Statement 1

yz is divisible by 4.

This means 'y' and 'z' together contain at least 2 factors of 2. It could possibly contain 3 or more, so we really don't know about what values X can take.

Statement 2

X,Y,Z are all not divisible by 4. (We have 3 2's to place but X,Y and Z each cannot take more than one 2)

This means X,Y and Z all have exactly only one factor of 2. Hence all 3 X,Y and Z will be even!! - Sufficient.
Top
Master | Next Rank: 500 Posts
Posts: 385
Joined: Fri Sep 23, 2011 9:02 pm
Thanked: 62 times
Followed by:6 members

by user123321 » Wed Nov 16, 2011 4:11 am
gunjan1208 wrote:If X, Y and z are integers and xyz is divisible by 8, isx even?

(1) yz is divisible by 4
(2) X, Y, and z are all NOTdivisible by4

OA after sometime?
1) yz is divisible by 4=>
since they gave yz is just div by 4, we dont know anything about its individual factors.
=> if yz is div by 8 then x is not req to be an even
=> if yz is div by just 4 then x should be an even.
hence insufficient.

2) x,y,z are all not div by 4(misunderstood this one :) )
so each of them should be div by 2 inorder for xyz to be div by 8.
hence sufficient.


+1 for B

user123321
Just started my preparation :D
Want to do it right the first time.
Top
Post new topic Post Reply
• Page 1 of 1