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For positive integer k, is the expression (k +

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For positive integer k, is the expression (k + [#permalink] New post 16 Sep 2008, 06:38
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For positive integer k, is the expression \((k + 2)(k^2 + 4k + 3)\) divisible by 4?

(1) k is divisible by 8.

(2) (k + 1)/3 is an odd integer.
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Re: Zumit DS 028 [#permalink] New post 16 Sep 2008, 06:55
dancinggeometry wrote:
For positive integer k, is the expression \((k + 2)(k^2 + 4k + 3)\) divisible by 4?

(1) k is divisible by 8.

(2) (k + 1)/3 is an odd integer.


from one:

(k+2) = even = 2(m+1) where m is devisible by 4..........a

(k^2 + 4k + 3) = even + even+ odd = odd..........b

thus from a & b

th expression boils down to

2*odd*odd thus we can deduct that k is not devisible by 4........ie: 1 is suff

from two

k+1 = odd and thus k is even could be ( check k = 0,2,4) we get multiple answers thus my answer is A
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Re: Zumit DS 028 [#permalink] New post 16 Sep 2008, 16:21
A for me as well


2 statement only tells that K is even
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Re: Zumit DS 028 [#permalink] New post 16 Sep 2008, 19:43
yezz wrote:
dancinggeometry wrote:
For positive integer k, is the expression \((k + 2)(k^2 + 4k + 3)\) divisible by 4?

(1) k is divisible by 8.

(2) (k + 1)/3 is an odd integer.


from one:

(k+2) = even = 2(m+1) where m is devisible by 4..........a

(k^2 + 4k + 3) = even + even+ odd = odd..........b

thus from a & b

th expression boils down to

2*odd*odd thus we can deduct that k is not devisible by 4........ie: 1 is suff

from two

k+1 = odd and thus k is even could be ( check k = 0,2,4) we get multiple answers thus my answer is A



Great solution/explanation, but you forgot that k is a positive integer.

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Re: Zumit DS 028 [#permalink] New post 16 Sep 2008, 22:30
(k+2)(k^2+4k+3) / 4 ***3 is the key...

1) Sufficient because with 8/k, k must be even. therefore, the equation is definitely NOT divisible by 4.

2) Sufficient. Work backwards with (k+1)/3=odd.

--> odd x 3 = odd.
--> k+1 = odd
--> k must be even.
--> if K is even, the equation is definitely NOT divisible by 4, so the final answer is D.
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Re: Zumit DS 028 [#permalink] New post 18 Sep 2008, 00:36
d2touge wrote:

2) Sufficient. Work backwards with (k+1)/3=odd.

--> odd x 3 = odd.
--> k+1 = odd
--> k must be even.
--> if K is even, the equation is definitely NOT divisible by 4, so the final answer is D.


Do not agree with this.

If (k+1)/3 = 1 then original expression will be 3*4*5 and will be divisible by 4.

But, if (k+1)/3 = 3 then original expression will be 9*10*11 and will not be divisible by 4.

Hence, stmt 2 is insufficient and answer should be A.
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Re: Zumit DS 028 [#permalink] New post 19 Sep 2008, 12:58
I agree with A as well ... from stat 2, the only thing I was able to derive was that k is even ...
Re: Zumit DS 028   [#permalink] 19 Sep 2008, 12:58
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