ds

This topic has expert replies
User avatar
Master | Next Rank: 500 Posts
Posts: 233
Joined: Mon Jun 09, 2008 1:30 am
Thanked: 5 times

ds

by blaster » Tue Oct 05, 2010 3:48 am
If x, y and z are different integers, is x divisible by 11?

1. xyz is divisible by 22 and 33
2. yz is divisible by 72
Source: — Data Sufficiency |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pm
Location: Milpitas, CA
Thanked: 447 times
Followed by:88 members

by Rahul@gurome » Tue Oct 05, 2010 4:23 am
(1) 22 = 2 * 11 and 33 = 3 * 11. Since xyz is divisible by 22 and 33, so xyz has the factors 2, 3 and 11. But this info is NOT SUFFICIENT to answer the question.

(2) 72 = 2^3 * 3^2, so yz has the factors 2 and 3. But this info is again NOT SUFFICIENT to answer the question.

Combining (1) and (2), also we cannot say whether x is divisible by 11.

[spoiler]The correct answer is (E).[/spoiler]
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Tue Oct 05, 2010 4:25 am
Is it E,
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.

User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 am
Location: Bangalore, India
Thanked: 146 times
Followed by:24 members

by shovan85 » Tue Oct 05, 2010 4:37 am
from 1: xyz multiple of 22 and 33. So we can say xyz = 66k (where k = 1, 2, 3, ...)
66 = LCM(22,33). But we cannot say whether x is a multiple of 11. Not Suff.

from 2: yz is a multiple of 72. So we can say yz=72k' (k'=1,2,3,....)
Still Not sufficient.

Combine 1 and 2 :

xyz = x(yz) = 66K
=> x(72k') = 66k
=> x(12k') = 11k
=> x = 11k/12k' Hence Not Suff.

IMO

E

Senior | Next Rank: 100 Posts
Posts: 77
Joined: Sat Sep 25, 2010 9:45 pm

by Yanat » Tue Oct 05, 2010 8:23 am
IMO it is E