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Problem Solving - Q

by gettingstarted » Thu Jul 21, 2011 11:10 am

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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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by GMATGuruNY » Thu Jul 21, 2011 11:44 am
gettingstarted wrote: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)
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.

The correct answer is A.

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

Answer choice A includes n(n+1).
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.
Last edited by GMATGuruNY on Sat Sep 10, 2011 4:48 am, edited 2 times in total.
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by gettingstarted » Fri Jul 22, 2011 10:10 am
Thanks for the reply. Its was really helpful.
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by leumas » 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.

The correct answer is A.

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

Answer choice A includes n(n+1).
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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by GMATGuruNY » 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.

The correct answer is A.

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

Answer choice A includes n(n+1).
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.
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].
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by [email protected] » Sun Jun 10, 2018 4:59 pm
Hi All,

We're told that N is an integer greater than 6. We're asked which of the following MUST be divisible by 3. This question can be solved rather easily by TESTing VALUES (as Mitch has shown). There's also a Number Property 'shortcut' built into this question that you can take advantage of.

To start, if we have 3 consecutive integers, then one of them MUST be divisible by 3. For example:
0, 1 and 2.... 0 is divisible by 3
1, 2 and 3... 3 is divisible by 3
2, 3 and 4... 3 is divisible by 3
3, 4 and 5... 3 is divisible by 3
4, 5 and 6... 6 is divisible by 3
Etc.

Within knowing the exact numbers, we can write them Algebraically as (N), (N+1) and (N+2). Increasing or decreasing any of those values by a 'multiple of 3' will still keep that number's respective "place" in line though (and it will continue to either be divisible by 3 or not). For example:

1, 2 and 3... if we increase the first term by 3, we end up with 4, 2 and 3 and the first term is still NOT divisible by 3 (it's '1 more' than a multiple of 3)
1, 2 and 3... if we increase the second term by 3, we end up with 2, 5 and 3 and the second term is still NOT divisible by 3 (it's '2 more' than a multiple of 3).
1, 2 and 3... if we increase the third term by 3, we end up with 1, 2 and 6 and the third term is STILL divisible by 3.

Thus, a sequence such as (N), (N+1) and (N+5) will include one term that is divisible by 3. In that same way, we could also subtract 3 from a term....
(N), (N-2) and (N+2) will include one term that is divisible by 3.

The correct answer will be the one that can be will give us 3 consecutive terms when we add 3s or subtract 3s from the terms.

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)

The third term in Answer A can be rewritten as (N+2) if we 'add 3' and then 'add 3' again. None of the other 4 Answers can be rewritten in this way.

Final Answer: A

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