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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gettingstarted
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We can plug in numbers and eliminate any answer that is not a multiple of 3.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)
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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gettingstarted
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Hi Gmatguru Thanks!! Now I got Anurag that n-4 can be written as -3+ (N-1),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.
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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GMATGuruNY
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When we plug in, we must comply with whatever conditions are given in the problem.leumas wrote:Hi Gmatguru Thanks!! Now I got Anurag that n-4 can be written as -3+ (N-1),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.
Then we should ignore the info given in the question that the integer is greater than 6.
Is this right approach?
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.
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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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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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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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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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
