prime & divisibility

This topic has expert replies
User avatar
Legendary Member
Posts: 659
Joined: Mon Dec 14, 2009 8:12 am
Thanked: 32 times
Followed by:3 members

prime & divisibility

by Gurpinder » Tue Aug 10, 2010 9:21 am
If n is an integer greater than 6, which of the following
must be divisible by 3 ?
(A) n(n + 1)(n - 4)
(B) n(n + 2)(n - 1)
(C) n(n + 3)(n - 5)
(D) n(n + 4)(n - 2)
(E) n(n + 5)(n - 6)

OA: A

please explain how you would handle such a problem
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
Source: — Problem Solving |

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Tue Aug 10, 2010 10:08 am
Gurpinder wrote:If n is an integer greater than 6, which of the following
must be divisible by 3 ?
(A) n(n + 1)(n - 4)
(B) n(n + 2)(n - 1)
(C) n(n + 3)(n - 5)
(D) n(n + 4)(n - 2)
(E) n(n + 5)(n - 6)

OA: A

please explain how you would handle such a problem
Plug values into each answer choice. Eliminate any answer choice that doesn't give us a factor that is a multiple of 3.

Plug in n=7:

A) 7*8*3. Since 3 is multiple of 3, hold onto A.
B) 7*9*6. Since 9 and 6 are each a multiple of 3, hold onto B.
C) 7*10*2. Since neither 7 nor 10 nor 2 is a multiple of 3, eliminate C.
D) 7*11*5. Since neither 7 nor 11 nor 5 is a multiple of 3, eliminate D.
E) 7*12*1. Since 12 is a multiple of 3, hold onto E.

Now plug n=8 into A, B and E only:

A) 8*9*4. Since 9 is a multiple of 3, hold onto A.
B) 8*10*7. Since neither 8 nor 10 nor 7 is a multiple of 3, eliminate B.
E) 8*13*2. Since neither 8 nor 13 nor 2 is a multiple of 3, eliminate E.

The correct answer is A.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

User avatar
Junior | Next Rank: 30 Posts
Posts: 20
Joined: Sun Aug 08, 2010 8:36 pm

by mainhoon » Tue Aug 10, 2010 9:07 pm
But how can one prove in general?

Senior | Next Rank: 100 Posts
Posts: 37
Joined: Thu Aug 05, 2010 3:15 am

by aarati » Tue Aug 10, 2010 10:39 pm
GMATGuruNY wrote: Plug values into each answer choice. Eliminate any answer choice that doesn't give us a factor that is a multiple of 3.

Plug in n=7:

A) 7*8*3. Since 3 is multiple of 3, hold onto A.
B) 7*9*6. Since 9 and 6 are each a multiple of 3, hold onto B.
C) 7*10*2. Since neither 7 nor 10 nor 2 is a multiple of 3, eliminate C.
D) 7*11*5. Since neither 7 nor 11 nor 5 is a multiple of 3, eliminate D.
E) 7*12*1. Since 12 is a multiple of 3, hold onto E.

Now plug n=8 into A, B and E only:

A) 8*9*4. Since 9 is a multiple of 3, hold onto A.
B) 8*10*7. Since neither 8 nor 10 nor 7 is a multiple of 3, eliminate B.
E) 8*13*2. Since neither 8 nor 13 nor 2 is a multiple of 3, eliminate E.

The correct answer is A.
this is the only way to solve this type of problem.... can you explain me clearly.....