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If n is an integer greater than 6 + divisibility question

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cramya Legendary Member Default Avatar
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If n is an integer greater than 6 + divisibility question

Post Sun Sep 14, 2008 5:16 pm
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.



Last edited by cramya on Mon Nov 24, 2008 4:21 am; edited 1 time in total

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alescau Junior | Next Rank: 30 Posts Default Avatar
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Post Sun Sep 14, 2008 5:55 pm
ok, in general every third number in sequence is divisible by three. by taking n, n+1, n+2 you are sure that one of them is divisible by 3. you can add or subtract 3 from either of them without changing divisibility.
n is the same as n-6, n-3, n+3, n+6...
n+1 is the same as n-5, n-2, n+4, n+7...
n+2 is the same as n-4, n-1, n+5, n+8...

now you must pick an answer that has one expression from each of these groups

__A__

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gmataug08 Master | Next Rank: 500 Posts Default Avatar
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Post Sun Sep 14, 2008 6:07 pm
IMO A

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4meonly Legendary Member Default Avatar
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Post Tue Sep 16, 2008 7:52 am
Obviously A

n (n+1) (n-4)
if you add to (n-4) 3 you'll get
(n-1) n (n+1)
3 consequtive integers. thier product is always divisible by 3

you can imagine it on number line

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Fab Senior | Next Rank: 100 Posts Default Avatar
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Post Sun Nov 23, 2008 3:59 pm
Could anyone elaborate more please?

Thanks.

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4meonly Legendary Member Default Avatar
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Post Mon Nov 24, 2008 12:27 am
Fab wrote:
Could anyone elaborate more please?

Thanks.
U can draw a number line for all given answers

(n-4)-(n-3)-(n-2)-(n-1)-(n)-(n+1)-------->

let n-4 = 1, than n=5, n=6, 1*5*6 is divisible by 3
any number for n-4 will yield a product od (n-4)*(n)*(n+1) that is divisible by 3

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mals24 Legendary Member Default Avatar
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Post Mon Nov 24, 2008 3:54 am
@ Fab

You can also plug in 7 and 8 as 'n' in each of the options. Only option A will work.

Cramya where did you hunt this question down from???

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cramya Legendary Member Default Avatar
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Post Mon Nov 24, 2008 4:22 am
Mals24,
It was sometime back and I dont tremember the source. I agree with all posters above that it should be A) and not D)

May be its a sets question and sometime the answers given cannot be trusted. I will try to loook for the source though if there is any way I could.

Regards,
Cramya

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Post Mon Nov 24, 2008 4:51 am
first of all, you must understand this fact: every nth integer is divisible by n.
in other words, every other integer is divisible by 2; every third integer is divisible by 3; etc.
make sure you understand this basic principle first; if you don't, then essentially none of the following will make any sense to you.

for problems like this one, in which you're considering the divisibility of unknown numbers, it can help to do the following:
(1) set up a number line, with ..., n - 1, n, n + 1, n + 2, ... placed along it
(2) write out the different CASES for which ones are divisible by 3 (or divisible by whatever number you're talking about; in this problem it just happens to be 3)

here's what i mean.

in this case, there are 3 different cases, because every third integer is divisible by 3. (in general, for divisibility by n, there will be n different ways in which to draw the number line.)
here are the 3 cases. in each case, the multiples of 3 are in orange, and the non-multiples of 3 are in the default color (black on my view, although i'm not sure whether that's the default for everyone else).

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 ...

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 (because, if there's a 3 anywhere in the factorization, then the whole product is a multiple of 3 - as with any other factor).

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.

note that the correct answer is a, not d.

--

note that you could also just plug in a ton of numbers and see what happens with the factorizations: just plug in 7, 8, 9, ... (per the directions) for the choices and see what the products are. don't multiply the products if you do that; just see whether the numbers in the products are divisible by 3.

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cramya Legendary Member Default Avatar
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Post Mon Nov 24, 2008 4:57 am
Thanks Ron for suggesting a nice approach and a detailed explanation!

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Fab Senior | Next Rank: 100 Posts Default Avatar
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Post Mon Nov 24, 2008 6:15 am
Quote:
@ Fab

You can also plug in 7 and 8 as 'n' in each of the options. Only option A will work.

If it's 8...C works...

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Post Mon Nov 24, 2008 6:42 am
Quote:
If it's 8...C works...
But 7 doesn't work. I said plug in 7 AND 8, and not 7 OR 8.

It should work for both of them not either of them.

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Fab Senior | Next Rank: 100 Posts Default Avatar
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Post Mon Nov 24, 2008 6:53 am
Got it.

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alina_linlin Newbie | Next Rank: 10 Posts Default Avatar
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Post Fri Feb 13, 2009 11:17 pm
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.

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sureshbala Master | Next Rank: 500 Posts
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Post Sat Feb 14, 2009 4:32 am
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.

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