1. If x^2 is divisible by 216 ,what is the smallest possible value for positive integer x?
I just have the explanation and the OA and no answer choices.
I will post the OA after some discussion.
Divisibility
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 298
- Joined: Tue Feb 16, 2010 1:09 am
- Thanked: 2 times
- Followed by:1 members
-
- Master | Next Rank: 500 Posts
- Posts: 437
- Joined: Sat Nov 22, 2008 5:06 am
- Location: India
- Thanked: 50 times
- Followed by:1 members
- GMAT Score:580
-
- Master | Next Rank: 500 Posts
- Posts: 298
- Joined: Tue Feb 16, 2010 1:09 am
- Thanked: 2 times
- Followed by:1 members
GMAT/MBA Expert
- Rahul@gurome
- GMAT Instructor
- Posts: 1179
- Joined: Sun Apr 11, 2010 9:07 pm
- Location: Milpitas, CA
- Thanked: 447 times
- Followed by:88 members
For smallest possible positive integer value of x, x² will be smallest too. Thus, we have to find smallest possible x² such that it is divisible by 216 (= 6³). Say, x² = 6³n. In other words, we have to find minimum possible positive integer value of n for which 6³n becomes a perfect square. Clearly n = 6 => x² = 6^4 => x = 6² = 36.Deepthi Subbu wrote:1. If x^2 is divisible by 216 ,what is the smallest possible value for positive integer x?
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)
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)