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# n and m are positive integers, can n divisible by 3? 1/ n

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n and m are positive integers, can n divisible by 3? 1/ n [#permalink]

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24 Sep 2009, 13:58
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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 ???
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Re: Divisible question [#permalink]

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24 Sep 2009, 18:18
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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.
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Re: Divisible question [#permalink]

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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.
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Re: Divisible question [#permalink]

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25 Sep 2009, 08:11
hi AKProdigy87 , good way of solving, but is this kind of substitution correct?
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Re: Divisible question [#permalink]

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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.
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Re: Divisible question [#permalink]

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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
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# n and m are positive integers, can n divisible by 3? 1/ n

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