Problem Solving

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 ?

rockeyb wrote: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 ?

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

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 .

kstv wrote: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.

as in many other problems, the "textbook" way to solve this problem is not the fastest way.
more to the point,
takeaway:
on problems involving divisibility by a particular number -- if you don't comprehend the theory of the problem RIGHT AWAY -- you should start making lists of numbers.

ALL problems involving divisibility depend on REPEATING PATTERNS, since divisibility is, at heart, little more than a pattern that repeats over and over again. (for instance, divisibility by 3 is just a pattern that repeats every three integers.)

so let's just make lists of numbers that satisfy each answer choice. we'll start with 7, since the problem stipulates that n must be greater than 6.

(a)
(7)(8)(3)
(8)(9)(4)
(9)(10)(5)
(10)(11)(6)
(11)(12)(7)
(12)(13)(8)
it looks like all of these are going to work, and there's a clear pattern. therefore, we actually don't even have to investigate the other choices, unless we have excess time.

IF we have the time to investigate the other choices, here is what we'll find:

(b)
(7)(9)(6)
(8)(10)(7) -- isn't divisible by 3

(c)
(7)(10)(2) -- isn't divisible by 3

(d)
(7)(11)(5) -- isn't divisible by 3

(e)
(7)(12)(1)
(8)(13)(2) -- isn't divisible by 3

--

(a) for the win.

--

the lesson you should get here:
if you have a problem on a topic that relates to REPEATING PATTERNS -- remainders, divisibilty, digits, etc. -- then you should START MAKING LISTS RIGHT AWAY if you don't know what to do.

@ rockeyb

rockeyb wrote: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.

the problem statement says:
which of the following must be divisible by 3?

"must" means "in ALL cases".
if you can find any instance in which the condition is not true, then the statement involving "must" becomes false.

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

this isn't worth worrying about.
the point is getting TAKEAWAYS that will apply to other problems -- these takeaways are unaffected by "difficulty level".
i.e., you can learn things from "easy" problems that will enable you to solve "hard" problems, and vice versa -- the point is that you note the similarities from one problem to the next.

by thinking about "difficulty levels", you're actually hindering your ability to learn takeaways that will solve other problems, because you'll start thinking about problems with "difficulty level X" as totally separate from problems with "difficulty level Y". that is exactly the sort of thinking that will neutralize the value of your study.

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

the fastest way to solve the problem is just to start making lists, as shown above. you can very easily do that in less than 1 minute.

i've had several students who refuse to accept the legitimacy of these alternative methods (making lists, plugging in numbers, etc.), and instead insist on trying to find algebraic solutions to every problem in the world. that's an inflexible viewpoint, and is toxic if you're looking to score at the highest levels.

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 .

rockeyb wrote: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 .

sure.

actually, it's more extreme than this -- you actually shouldn't think at all about difficulty levels, for at least four really good reasons:

(1) as stated above, they will cause you to think of problems of different "difficulty levels" as unrelated; this is the exact opposite of the optimal approach (which is to look for connections between ANY problems, ANYWHERE).

(2) you won't know "difficulty levels" on the actual test.
therefore, the "difficulty levels" of practice problems will either be irrelevant to you or actually hurt you. (i.e., if you devise actual test-taking strategies based on "difficulty levels", then this will actually hurt you on test day -- you'll be trying to base strategies on information that you don't have!)

(3) optimal time management strategy is independent of "difficulty level".
you should be spending the same allotment of time on each question. in fact, if a problem is clearly too hard for you, you should be spending LESS time on it!
when people worry about "difficulty level", they tend to spend more time on problems they perceive as harder. this strategy will be detrimental to your success.

(4) it increases your stress level, in exchange for no benefit whatsoever.

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