og 82

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.

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?

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

expression given in the options will be divisible by 3, if any of the term in the expression is divisible by 3,

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

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)

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

Picking numbers is a great approach and, using a bit of common sense, you should realize that you only have to try two numbers.

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).

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

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

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.

thnks stuart og solved it trial and error basis bt i thnk it could be time consuming....thnks stuart

Stuart Kovinsky wrote:

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

Picking numbers is a great approach and, using a bit of common sense, you should realize that you only have to try two numbers.

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).