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)
how to solve ths problem other than the way of og
og 82
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- Bhanu Theja
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Suppose you group the positive integers as below.
1-A
2-B
3-C
4-A
5-B
6-C
..
..
..
If you multiply one number each from groups A, B and C then the result will be divisible by three.
Indexing as ABC can be started from any number.
That is so because you are dividing numbers into 3 groups and one of the groups will have all the factors of 3.
Now you have a number 'n'
Lets say
(n-6) - A
(n-5) - B
(n-4) - C
(n-3) - A
(n-2) - B
(n-1) - C
n - A
(n+1) - B
(n+2) - C
(n+3) - A
(n+4) - B
(n+5) - C
Now look up for options which have combination of A, B and C.
Only (A)
This is not a GENIUS' way to solve it but you asked for it
P.S - If they ask for a number divisible by 4. Then you can group them as A, B, C and D.
1-A
2-B
3-C
4-A
5-B
6-C
..
..
..
If you multiply one number each from groups A, B and C then the result will be divisible by three.
Indexing as ABC can be started from any number.
That is so because you are dividing numbers into 3 groups and one of the groups will have all the factors of 3.
Now you have a number 'n'
Lets say
(n-6) - A
(n-5) - B
(n-4) - C
(n-3) - A
(n-2) - B
(n-1) - C
n - A
(n+1) - B
(n+2) - C
(n+3) - A
(n+4) - B
(n+5) - C
Now look up for options which have combination of A, B and C.
Only (A)
This is not a GENIUS' way to solve it but you asked for it
P.S - If they ask for a number divisible by 4. Then you can group them as A, B, C and D.
Last edited by Bhanu Theja on Wed Apr 20, 2011 6:52 am, edited 1 time in total.
Bhanu Theja
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Hi,
I think the first suggested solution in the OG is quite good for solving this problem. You choose 7 first and can eliminate C and D and then by choosing the next bigger number (which here is 8) you test the other 3 possibilites and get an answer.
Why are you looking for another solution? Do you think this solution takes too much time?
I think the first suggested solution in the OG is quite good for solving this problem. You choose 7 first and can eliminate C and D and then by choosing the next bigger number (which here is 8) you test the other 3 possibilites and get an answer.
Why are you looking for another solution? Do you think this solution takes too much time?
- manpsingh87
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expression given in the options will be divisible by 3, if any of the term in the expression is divisible by 3,nafiul9090 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)
how to solve ths problem other than the way of og
now consider only those numbers which are not multiples of 3, because all other numbers which are multiples of 3 will easily divide any of the given option.
numbers which are not multiples of 3 will be of the type 3k+1 or 3k+2, i.e. these numbers will leave reminder 1 and 2 when divided by 3 respectively.
now first of put n=3k+1, in the first option,
n(n+1)(n-4);(3k+1)*(3k+1+1)*(3k+1-4);(3k+1)*(3k+2)*(3k-3)= 3{(3k+1)*(3k+2)*(k-1)}; hence expression will be divisible by 3.
now put n=3k+2
n(n+1)(n-4); (3k+2)*(3k+2+1)*(3k+2-4); (3k+2)*(3k+3)*(3k-2);3{(3k+2)*(k+1)*(3k-2)}; hence expression will be divisible by 3.
as option 1 is divisible for all numbers which leaves remainder 1 or 2 when divided by 3, therefore answer should be A
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Though i have still not yet started with my og prep, i am gng thrugh few other resources at present..!!
Saw the post and just thught of attempting this...
so here it goes, its easy...
pick numbers,
i picked 7 and 20 and quickly substitute only A will be divisible by 3 for both the substitutions
--
Never Give up, Keep on fighting (borrowed quote)
Saw the post and just thught of attempting this...
so here it goes, its easy...
pick numbers,
i picked 7 and 20 and quickly substitute only A will be divisible by 3 for both the substitutions
--
Never Give up, Keep on fighting (borrowed quote)
- Stuart@KaplanGMAT
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Picking numbers is a great approach and, using a bit of common sense, you should realize that you only have to try two numbers.nafiul9090 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)
how to solve ths problem other than the way of og
First, we recognize that if n is a multiple of 3 then every choice will also be a multiple of 3.
Second, since "3" is the important number, we recognize that there are only two other possible remainders when you divide by 3: 1 and 2. So, we just need to pick one number that's a multiple of 3 plus 1 and one number that's a multiple of 3 plus 2 to cover every possibility.
Since n>6 and the biggest "minus" term is (n-6), plugging in n=7 and n=8 sounds like a great plan.
We can solve without picking numbers by realizing that if any of the 3 terms in a choice is a multiple of 3, the entire expression will be a multiple of 3. Since there are 3 terms, to guarantee that the whole expression is a multiple of 3 we need the terms to be in the form (k)(l+1)(m+2) in which k, l and m are all multiples of 3. The correct answer will fit this pattern.
Only (A) fits the bill, since (n-4) plays the same role as (n+2).
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- manpsingh87
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hi,its incorrect, a number will be divisible by 3 if sum of its digits is divisible by 3, for example 3*4=12, which is divisible by because sum of its digit=1+2=3 is divisible by 3, but 3+4 =7 is not divisible by 3.ces444ces wrote:Another way is applying the rule that any integer is divisible by 3 if the sum of the terms of the integer is divisible by 3.
Taking a look at (A):
n + (n+1) + (n-4) = 3n-3 = 3(n-1) <---always a multiple of 3
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thnks stuart og solved it trial and error basis bt i thnk it could be time consuming....thnks stuart
Stuart Kovinsky wrote:Picking numbers is a great approach and, using a bit of common sense, you should realize that you only have to try two numbers.nafiul9090 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)
how to solve ths problem other than the way of og
First, we recognize that if n is a multiple of 3 then every choice will also be a multiple of 3.
Second, since "3" is the important number, we recognize that there are only two other possible remainders when you divide by 3: 1 and 2. So, we just need to pick one number that's a multiple of 3 plus 1 and one number that's a multiple of 3 plus 2 to cover every possibility.
Since n>6 and the biggest "minus" term is (n-6), plugging in n=7 and n=8 sounds like a great plan.
We can solve without picking numbers by realizing that if any of the 3 terms in a choice is a multiple of 3, the entire expression will be a multiple of 3. Since there are 3 terms, to guarantee that the whole expression is a multiple of 3 we need the terms to be in the form (k)(l+1)(m+2) in which k, l and m are all multiples of 3. The correct answer will fit this pattern.
Only (A) fits the bill, since (n-4) plays the same role as (n+2).