|
Author |
Message |
|
TAGS:
|
|
|
Intern
Joined: 24 Sep 2009
Posts: 17
Followers: 0
Kudos [?]:
4
[0], given: 0
|
n and m are positive integers, can n divisible by 3? 1/ n [#permalink]
 24 Sep 2009, 13:58
Question Stats:
100% (04:33) correct
0% (00:00) wrong based on 0 sessions
n and m are positive integers, can n divisible by 3?
1/ n divisible by m(m^2+2)
2/ n divisible by m^2(m+2)
I guess it's C ???
|
|
|
|
|
|
|
Manager
Joined: 11 Sep 2009
Posts: 129
Followers: 3
Kudos [?]:
87
[1] , given: 6
|
Re: Divisible question [#permalink]
24 Sep 2009, 18:18
1
This post received
KUDOS
The answer is A.
Any integer can be represented in terms of divisibility by 3 by placing them in the following 3 groups:
a) (3x) - Divisible by 3 (ex. 3, 6, 9, etc.)
b) (3x + 1) - Remainder of 1 (ex. 4, 7, 10, etc.)
c) (3x + 2) - Remainder of 2 (ex. 5, 8, 11, etc.)
where x is an integer.
Statement 1: n divisible by m(m^2+2):
a) Let m = 3x
m(m^2 + 2) = (3x)((3x)^2 + 2) = 3(x)(9x^2 + 2) DIVISIBLE BY 3
b) Let m = 3x + 1
m(m^2 + 2) = (3x+1)((3x+1)^2 + 2) = (3x + 1)(9x^2 + 6x + 1 + 2)
= (3x + 1)(9x^2 + 6x + 3) = 3(3x^2 + 2x + 1)(3x + 1) DIVISIBLE BY 3
c) Let m = 3x + 2
m(m^2 + 2) = (3x+2)((3x+2)^2 + 2) = (3x + 1)(9x^2 + 12x + 4 + 2)
= (3x + 1)(9x^2 + 12x + 6) = 3(3x^2 + 4x + 2)(3x + 1) DIVISIBLE BY 3
Therefore Statement 1 is sufficient. If n is divisible by m(m^2 + 2), then it is divisible by 3.
Statement 2: n divisible by m^2(m+2):
...
c) Let m = 3x + 2
[m]m^2(m + 2) = (3x+2)(3x+2)(3x+2+2) = (3x+2)(3x+2)(3x+4) NOT DIVISIBLE BY 3
Therefore Statement 2 is not sufficient.
The answer is A.
|
|
|
|
|
|
Intern
Joined: 24 Sep 2009
Posts: 17
Followers: 0
Kudos [?]:
4
[0], given: 0
|
Re: Divisible question [#permalink]
24 Sep 2009, 20:29
AKProdigy87 wrote: The answer is A.
Any integer can be represented in terms of divisibility by 3 by placing them in the following 3 groups:
a) (3x) - Divisible by 3 (ex. 3, 6, 9, etc.)
b) (3x + 1) - Remainder of 1 (ex. 4, 7, 10, etc.)
c) (3x + 2) - Remainder of 2 (ex. 5, 8, 11, etc.)
where x is an integer.
Statement 1: n divisible by m(m^2+2):
a) Let m = 3x
m(m^2 + 2) = (3x)((3x)^2 + 2) = 3(x)(9x^2 + 2) DIVISIBLE BY 3
b) Let m = 3x + 1
m(m^2 + 2) = (3x+1)((3x+1)^2 + 2) = (3x + 1)(9x^2 + 6x + 1 + 2)
= (3x + 1)(9x^2 + 6x + 3) = 3(3x^2 + 2x + 1)(3x + 1) DIVISIBLE BY 3
c) Let m = 3x + 2
m(m^2 + 2) = (3x+2)((3x+2)^2 + 2) = (3x + 1)(9x^2 + 12x + 4 + 2)
= (3x + 1)(9x^2 + 12x + 6) = 3(3x^2 + 4x + 2)(3x + 1) DIVISIBLE BY 3
Therefore Statement 1 is sufficient. If n is divisible by m(m^2 + 2), then it is divisible by 3.
Statement 2: n divisible by m^2(m+2):
...
c) Let m = 3x + 2
[m]m^2(m + 2) = (3x+2)(3x+2)(3x+2+2) = (3x+2)(3x+2)(3x+4) NOT DIVISIBLE BY 3
Therefore Statement 2 is not sufficient.
The answer is A. Hi AKProdigy87, I don't quite understand that. Say we have n=4 and m=2. Then m(m^2+2)=2(2^2+2)=12 n=4 is devisible by 12 but doesn't mean 3 is divisible by n.
|
|
|
|
|
|
Manager
Joined: 14 Dec 2008
Posts: 179
Followers: 1
Kudos [?]:
12
[0], given: 39
|
Re: Divisible question [#permalink]
25 Sep 2009, 08:11
hi AKProdigy87 , good way of solving, but is this kind of substitution correct?
|
|
|
|
|
|
Manager
Joined: 08 Jul 2009
Posts: 178
Followers: 3
Kudos [?]:
12
[0], given: 13
|
Re: Divisible question [#permalink]
25 Sep 2009, 11:34
swat wrote: AKProdigy87 wrote: The answer is A.
Any integer can be represented in terms of divisibility by 3 by placing them in the following 3 groups:
a) (3x) - Divisible by 3 (ex. 3, 6, 9, etc.)
b) (3x + 1) - Remainder of 1 (ex. 4, 7, 10, etc.)
c) (3x + 2) - Remainder of 2 (ex. 5, 8, 11, etc.)
where x is an integer.
Statement 1: n divisible by m(m^2+2):
a) Let m = 3x
m(m^2 + 2) = (3x)((3x)^2 + 2) = 3(x)(9x^2 + 2) DIVISIBLE BY 3
b) Let m = 3x + 1
m(m^2 + 2) = (3x+1)((3x+1)^2 + 2) = (3x + 1)(9x^2 + 6x + 1 + 2)
= (3x + 1)(9x^2 + 6x + 3) = 3(3x^2 + 2x + 1)(3x + 1) DIVISIBLE BY 3
c) Let m = 3x + 2
m(m^2 + 2) = (3x+2)((3x+2)^2 + 2) = (3x + 1)(9x^2 + 12x + 4 + 2)
= (3x + 1)(9x^2 + 12x + 6) = 3(3x^2 + 4x + 2)(3x + 1) DIVISIBLE BY 3
Therefore Statement 1 is sufficient. If n is divisible by m(m^2 + 2), then it is divisible by 3.
Statement 2: n divisible by m^2(m+2):
...
c) Let m = 3x + 2
[m]m^2(m + 2) = (3x+2)(3x+2)(3x+2+2) = (3x+2)(3x+2)(3x+4) NOT DIVISIBLE BY 3
Therefore Statement 2 is not sufficient.
The answer is A. Hi AKProdigy87, I don't quite understand that. Say we have n=4 and m=2. Then m(m^2+2)=2(2^2+2)=12 n=4 is devisible by 12 but doesn't mean 3 is divisible by n. The q is saying n is divisible by m(m^2+2) if m is 12 then the minimum value of n is 12 which is divisible by 3.
|
|
|
|
|
|
Manager
Joined: 27 Oct 2008
Posts: 188
Followers: 1
Kudos [?]:
42
[0], given: 3
|
Re: Divisible question [#permalink]
25 Sep 2009, 23:39
Yes. Statement 1 alone is sufficient
taking m(m^2+2) and since m is positive integer, you can take m = 1,2,3 and so on
for any such value of m, this equation m(m^2+2) will always be a multiple of 3.
Since its a multiple of 3 and is a factor of n, thus n will be a multiple of 3 or in other words n will be divisible b 3.
Statement 2 is not sufficient
taking m^2(m+2) and since m is positive integer, you can take m = 1,2,3 and so on
for any such value of m, this equation m^2(m+2), seems to be a multiple of 3 in some cases and not a multiple in some cases. So based on this alone we cannot guess if n is divisible by 3 or not.
Hence A.
|
|
|
|
|
|
|
Re: Divisible question
[#permalink]
25 Sep 2009, 23:39
|
|
|
|
|
|
|
|
|
Similar topics |
Author |
Replies |
Last post |
|
Similar
Topics:
|
 |
|
|
If n is a positive integer, is n^3 - n divisible by 4? (1) n
|
TeHCM |
15 |
24 Oct 2005, 22:12 |
 |
|
|
If n is a positive integer, is n^3-n divisible by 4? 1) n =
|
Matador |
7 |
15 Apr 2006, 21:07 |
|
|
|
|
If n is a positive integer, is n^3 - n divisible by 4? (1) n
|
focused07 |
5 |
20 Jan 2007, 22:26 |
|
|
|
|
If n is a positive integer, is n^3 - n divisible by 4? 1) n
|
asaf |
7 |
27 Jul 2007, 21:55 |
|
|
|
|
If n is positive integer, is (n^3 - n) divisible by 4? 1. n
|
LoyalWater |
5 |
08 Sep 2008, 14:34 |
|
|
|
|
|
|
close
