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## Problem Solving - Q

This topic has 2 expert replies and 2 member replies
gettingstarted Newbie | Next Rank: 10 Posts
Joined
21 Jul 2011
Posted:
2 messages

#### Problem Solving - Q

Thu Jul 21, 2011 11:10 am
What is the best way to solve the Question below, is it by testing numbers or is there any other approach that can be tried, please suggest and explain the approach. Thanks

Q ) 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)

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### GMAT/MBA Expert

GMATGuruNY GMAT Instructor
Joined
25 May 2010
Posted:
13608 messages
Followed by:
1796 members
13060
GMAT Score:
790
Sat Sep 10, 2011 2:01 pm
leumas wrote:
GMATGuruNY wrote:
Q ) 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)

We can plug in numbers and eliminate any answer that is not a multiple of 3.

Let n=7:
A) n(n+1)(n-4) = 7*8*3, which is a multiple of 3. Hold onto A.
B) n(n+2)(n-1) = 7*9*6, which is a multiple of 3. Hold onto B.
C) n(n+3)(n-5) = 7*10*5, which is not a multiple of 3. Eliminate C.
D) n(n+4)(n-2) = 7*11*5, which is not a multiple of 3. Eliminate D.
E) n(n+5)(n-6) = 7*12*1, which is a multiple of 3. Hold onto E.

Let n=8:
A) n(n+1)(n-4) = 8*9*4, which is a multiple of 3. Hold onto A.
B) n(n+2)(n-1) = 8*10*7, which is not a multiple of 3. Eliminate B.
E) n(n+5)(n-6) = 8*13*2, which is not a multiple of 3. Eliminate E.

We also could reason our way to the correct answer.

Of every 3 consecutive integers, exactly one will be a multiple of 3:
1,2,3
2,3,4
3,4,5
4,5,6

Of n-1, n, and n+1 -- three consecutive integers -- exactly one will be a multiple of 3.
If either n or n+1 is a multiple of 3, then n(n+1) will be a multiple of 3.
If neither n nor n+1 is a multiple of 3, then n-1 must be a multiple of 3.
If n-1 is a multiple of 3, so is n-4, since the distance between them = (n-1) - (n-4) = 3.
If n-4 is a multiple of 3, then n(n+1)(n-4) is a multiple of 3.

Thus, whatever the scenario, answer choice A must be a multiple of 3.

I think plugging in numbers is MUCH easier.
Hi Gmatguru Thanks!! Now I got Anurag that n-4 can be written as -3+ (N-1),

Then we should ignore the info given in the question that the integer is greater than 6.
Is this right approach?
When we plug in, we must comply with whatever conditions are given in the problem.
In the problem here, n must be greater than 6.
In my solution, I plugged in n=7 and n=8, both of which are greater than 6.

_________________
Mitch Hunt
GMAT Private Tutor
GMATGuruNY@gmail.com
If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
Available for tutoring in NYC and long-distance.

Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.
leumas Senior | Next Rank: 100 Posts
Joined
21 Aug 2011
Posted:
44 messages
3
Test Date:
12/28/2011
Target GMAT Score:
800
GMAT Score:
NA
Sat Sep 10, 2011 5:20 am
GMATGuruNY wrote:
Q ) 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)

We can plug in numbers and eliminate any answer that is not a multiple of 3.

Let n=7:
A) n(n+1)(n-4) = 7*8*3, which is a multiple of 3. Hold onto A.
B) n(n+2)(n-1) = 7*9*6, which is a multiple of 3. Hold onto B.
C) n(n+3)(n-5) = 7*10*5, which is not a multiple of 3. Eliminate C.
D) n(n+4)(n-2) = 7*11*5, which is not a multiple of 3. Eliminate D.
E) n(n+5)(n-6) = 7*12*1, which is a multiple of 3. Hold onto E.

Let n=8:
A) n(n+1)(n-4) = 8*9*4, which is a multiple of 3. Hold onto A.
B) n(n+2)(n-1) = 8*10*7, which is not a multiple of 3. Eliminate B.
E) n(n+5)(n-6) = 8*13*2, which is not a multiple of 3. Eliminate E.

We also could reason our way to the correct answer.

Of every 3 consecutive integers, exactly one will be a multiple of 3:
1,2,3
2,3,4
3,4,5
4,5,6

Of n-1, n, and n+1 -- three consecutive integers -- exactly one will be a multiple of 3.
If either n or n+1 is a multiple of 3, then n(n+1) will be a multiple of 3.
If neither n nor n+1 is a multiple of 3, then n-1 must be a multiple of 3.
If n-1 is a multiple of 3, so is n-4, since the distance between them = (n-1) - (n-4) = 3.
If n-4 is a multiple of 3, then n(n+1)(n-4) is a multiple of 3.

Thus, whatever the scenario, answer choice A must be a multiple of 3.

I think plugging in numbers is MUCH easier.
Hi Gmatguru Thanks!! Now I got Anurag that n-4 can be written as -3+ (N-1),

Then we should ignore the info given in the question that the integer is greater than 6.
Is this right approach?

### GMAT/MBA Expert

GMATGuruNY GMAT Instructor
Joined
25 May 2010
Posted:
13608 messages
Followed by:
1796 members
13060
GMAT Score:
790
Sat Sep 10, 2011 2:01 pm
leumas wrote:
GMATGuruNY wrote:
Q ) 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)

We can plug in numbers and eliminate any answer that is not a multiple of 3.

Let n=7:
A) n(n+1)(n-4) = 7*8*3, which is a multiple of 3. Hold onto A.
B) n(n+2)(n-1) = 7*9*6, which is a multiple of 3. Hold onto B.
C) n(n+3)(n-5) = 7*10*5, which is not a multiple of 3. Eliminate C.
D) n(n+4)(n-2) = 7*11*5, which is not a multiple of 3. Eliminate D.
E) n(n+5)(n-6) = 7*12*1, which is a multiple of 3. Hold onto E.

Let n=8:
A) n(n+1)(n-4) = 8*9*4, which is a multiple of 3. Hold onto A.
B) n(n+2)(n-1) = 8*10*7, which is not a multiple of 3. Eliminate B.
E) n(n+5)(n-6) = 8*13*2, which is not a multiple of 3. Eliminate E.

We also could reason our way to the correct answer.

Of every 3 consecutive integers, exactly one will be a multiple of 3:
1,2,3
2,3,4
3,4,5
4,5,6

Of n-1, n, and n+1 -- three consecutive integers -- exactly one will be a multiple of 3.
If either n or n+1 is a multiple of 3, then n(n+1) will be a multiple of 3.
If neither n nor n+1 is a multiple of 3, then n-1 must be a multiple of 3.
If n-1 is a multiple of 3, so is n-4, since the distance between them = (n-1) - (n-4) = 3.
If n-4 is a multiple of 3, then n(n+1)(n-4) is a multiple of 3.

Thus, whatever the scenario, answer choice A must be a multiple of 3.

I think plugging in numbers is MUCH easier.
Hi Gmatguru Thanks!! Now I got Anurag that n-4 can be written as -3+ (N-1),

Then we should ignore the info given in the question that the integer is greater than 6.
Is this right approach?
When we plug in, we must comply with whatever conditions are given in the problem.
In the problem here, n must be greater than 6.
In my solution, I plugged in n=7 and n=8, both of which are greater than 6.

_________________
Mitch Hunt
GMAT Private Tutor
GMATGuruNY@gmail.com
If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
Available for tutoring in NYC and long-distance.

Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.
leumas Senior | Next Rank: 100 Posts
Joined
21 Aug 2011
Posted:
44 messages
3
Test Date:
12/28/2011
Target GMAT Score:
800
GMAT Score:
NA
Sat Sep 10, 2011 5:20 am
GMATGuruNY wrote:
Q ) 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)

We can plug in numbers and eliminate any answer that is not a multiple of 3.

Let n=7:
A) n(n+1)(n-4) = 7*8*3, which is a multiple of 3. Hold onto A.
B) n(n+2)(n-1) = 7*9*6, which is a multiple of 3. Hold onto B.
C) n(n+3)(n-5) = 7*10*5, which is not a multiple of 3. Eliminate C.
D) n(n+4)(n-2) = 7*11*5, which is not a multiple of 3. Eliminate D.
E) n(n+5)(n-6) = 7*12*1, which is a multiple of 3. Hold onto E.

Let n=8:
A) n(n+1)(n-4) = 8*9*4, which is a multiple of 3. Hold onto A.
B) n(n+2)(n-1) = 8*10*7, which is not a multiple of 3. Eliminate B.
E) n(n+5)(n-6) = 8*13*2, which is not a multiple of 3. Eliminate E.

We also could reason our way to the correct answer.

Of every 3 consecutive integers, exactly one will be a multiple of 3:
1,2,3
2,3,4
3,4,5
4,5,6

Of n-1, n, and n+1 -- three consecutive integers -- exactly one will be a multiple of 3.
If either n or n+1 is a multiple of 3, then n(n+1) will be a multiple of 3.
If neither n nor n+1 is a multiple of 3, then n-1 must be a multiple of 3.
If n-1 is a multiple of 3, so is n-4, since the distance between them = (n-1) - (n-4) = 3.
If n-4 is a multiple of 3, then n(n+1)(n-4) is a multiple of 3.

Thus, whatever the scenario, answer choice A must be a multiple of 3.

I think plugging in numbers is MUCH easier.
Hi Gmatguru Thanks!! Now I got Anurag that n-4 can be written as -3+ (N-1),

Then we should ignore the info given in the question that the integer is greater than 6.
Is this right approach?

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