Folks, this can be answered quickly in the following way.

When n is divided by 3, the possible remainders are 0, 1 and 2.

When the remainder is 0, obviously all the options will be divisible by 3.

If n leaves a remainder 1, only options A, B and E are divisible by 3.

If n leaves a remainder 2, out of options A, B and E only option A is divisible by 3.

Hence A

sureshbala wrote:

alina_linlin wrote:I just come out an idea that, whether any 3 intergers whose sum could be divided by 3, can also make a product divided by 3?

take the answer A for instance, n+n+1+n-4=3n-3, can be divided by 3.

for the other 4 answers, no one can meet the above rules.

Friend this is not true in all the cases....

If a+b+c is divisible by 3, you can't conclude that axbxc will always be divisible by 3.

Here is an example....

7+10+13 is divisible by 3, but 7x10x13 is not divisible by 3.

Similarly the converse is also not true...

i.e If axbxc is divisible by 3, we can conclude that a+b+c will be divisible by 3.

Here is an example....

9x10x13 is divisible by 3, but 9+10+13 is not divisible by 3.

Your example is solid, but is not applicable to alina's example or this problem.

Hi Ron ,

Now my question is why is A the correct answer and why not B , C or D .

Since the question asks which one is divisible by 3 , if the product of three numbers contain at least one multiple of 3 then that product is divisible by 3 right ?

So why are we neglecting the other options ?

am I missing some thing ?

Just trying to consolidate what has been discussed above.
Lunarpower every nth integer is divisible by n.
sureshbala : If a+b+c is divisible by 3, you can't conclude that axbxc will always be divisible by 3.
But we know if they are odd consecutive nos it is possible
https://www.beatthegmat.com/3-consecutiv ... 54101.html
as it is easier to deal with +sign then - sign
let us assume n = 6+k (n> 6) and arrange them in ascending order
A ) A. n (n+1) (n-4) can be written as (6+k) (7+k) (2+k)
Ascending order (2+k) (6+k) (7+k)

B. n (n+2) (n-1) after arranging (5+k) (6+k) (8+k)
C. n (n+3) (n-5) " (1+k) (6+k) (9+k)
D. n (n+4) (n-2) " (4+k) (6+k) (10+k)
E. n (n+5) (n-6) " (k) (6+k) (11+k)

If they were consecutive nos eg (x-1) (x) and (x+1) whose sum = 3x it will be divisible by 3
But if we replace (x+1) with (x+1+3) = (x+4) then also the sum of three nos (x-1) (x) and (x+4) = 3x + 3 is divisible by 3.
same for (x-1-3) (x) (x+1) = 3x-3 = 3(x-1) is divisible by 3
So we can replace the third consecutive no with addition or substraction of 3 as in the above case.

Only A and B have atleast two consecutive nos so most probable options

Take B - (5+k)(6+k) if the next no was (7+k) or (7+3+k) it could suffice but it is (k+8)

Consider A (6+k) (7+k) are consecutive nos
So (6+k-1) (6+k-4) should do the trick.
(6+k-4) is (2+k) , the first term

So option A, but how will you spot this in 2 mins.
Arranging them in ascending order will take more than 1 min . There is a way.

lunarpower wrote: hi --

did you read the following post?
https://www.beatthegmat.com/if-n-is-an-i ... html#99525

in that post, i specifically laid out situations in which each of those four choices lacks a multiple of 3.

if there are things in that post that you had trouble understanding, then i can certainly try to explain them further -- but i wanted to make sure that you'd actually read the post.

thanks

Yes i have seen this post and I will try to clarify my question a bit . I understand the fact that we are setting up 3 cases in order to find out if there is at least one multiple of 3 .

case 1

... n-6 n-5 n-4 n-3 n-2 n-1 n n+1 n+2 n+3 n+4 n+5 n+6 ...

case 2
... n-6 n-5 n-4 n-3 n-2 n-1 n n+1 n+2 n+3 n+4 n+5 n+6 ...

case 3
... n-6 n-5 n-4 n-3 n-2 n-1 n n+1 n+2 n+3 n+4 n+5 n+6 ...

Now here is what I dont understand why are we only focusing on case 3 why not other cases.

once you've done this, just look at the 3 cases for each of the answer choices. if any of the factors is in orange, then that particular product is a multiple of 3

for choice (a), all three cases include one orange factor apiece: n is orange in case 1, n+1 is orange in case 3, and n-4 is orange in case 2. therefore, (a) is the correct answer.

for choice (b), case 3 contains no orange factors, so (b) doesn't have to be a multiple of 3.

for choice (c), case 2 contains no orange factors, so (b) doesn't have to be a multiple of 3.

for choice (d), case 2 contains no orange factors, so (b) doesn't have to be a multiple of 3.

for choice (e), case 3 contains no orange factors, so (b) doesn't have to be a multiple of 3.

In the above choices only option A has at least one orange factor in all 3 cases .

But if we look at other choices they too have atleast one orange factor but not in all the 3 cases Ex : choice (b) has two orange factor in case 2 but none in 1 and 3 . So how dose this rule out that choice (b) is not a multiple of 3.

OR are we saying that choice(b) may be a multiple of 3 but not always , similarly for other options too .

Also what difficulty level is this problem for is it 600 -700 level ?

And I agree with kstv how can we solve this in 2 mins is there a shorter method ?

I guess I have posted a lot of questions in one post sorry for that .

Thanks a lot Ron for that explanation .

I agree worry too much about difficulty level will hinder my performance instead I should be focusing on the take away from this problem .

Thanks again for your help its greatly appreciated .

Yup agreed nice one , hope this helps others too .

cramya 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)

Answer is given as d Can someone pl explain?

If i take n = 7 then d will not work.

Hi cramya,

It's nice to find your post after a long time. I congratulate you for providing a small glitch (in the OA) in the question that resulted in a greatly useful explanation ultimately. Ron and suresh have been nice to do all justice with many great doubts from great members. For me, this was a purely Number Line problem and I have no doubt in calling [spoiler]A[/spoiler] as my answer. Good explanations are already there.

Keep posting great questions like this, please!!

That is the problem Q82 in the OG12.
The official answer is A, but explanation takes 3/5 of the page!!!
For me personally, the most elegant solution was to think about n-4, n-1, n+2 etc

sureshbala wrote:

alina_linlin wrote:I just come out an idea that, whether any 3 intergers whose sum could be divided by 3, can also make a product divided by 3?

take the answer A for instance, n+n+1+n-4=3n-3, can be divided by 3.

for the other 4 answers, no one can meet the above rules.

Friend this is not true in all the cases....

If a+b+c is divisible by 3, you can't conclude that axbxc will always be divisible by 3.

Here is an example....

7+10+13 is divisible by 3, but 7x10x13 is not divisible by 3.

Similarly the converse is also not true...

i.e If axbxc is divisible by 3, we can conclude that a+b+c will be divisible by 3.

Here is an example....

9x10x13 is divisible by 3, but 9+10+13 is not divisible by 3.

Alina's method is based on the simple principle that sum of n consecutive integers is always divisible by n if n is odd. This however does not hold true for even integers.

Thanks Ron for your explanation!

Source OG12 Q82 Page 163