MGMAT DS

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MGMAT DS

by PAB2706 » Thu Oct 23, 2008 4:49 am
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.

not understanding the approach to this problem

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by bharathaitha » Thu Oct 23, 2008 5:01 am
The expression (K+2)(K^2+4K+3) can be reduced to

(K+2)(K+1)(K+3) ==> (K+1)(K+2)(K+3)

Statement A

K/8 is an integer i.e. k= 8, 16, 24, 32, 40, 48 etc. for all those values (K+1)(K+2)(K+3) is not divisible by 4

Hence SUFFICIENT

Statement B

(K+1)/3 is an odd integer, then K = 2, 8, 14, 20

For K = 8, (K+1)(K+2)(K+3) is not divisible by 4
For K = 14, (K+1)(K+2)(K+3) is divisible by 4

Hence INSUFFICIENT

Therefore the OA is A

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Re: MGMAT DS

by drpawan » Thu Oct 23, 2008 5:16 am
PAB2706 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.

not understanding the approach to this problem
Same explanation as bharathaitha.
ans is A

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by stop@800 » Thu Oct 23, 2008 6:12 am
A different explanation but answer is A :)


A
K is divisible by 8
hence
K is divisible by 4
so
K+2 will not be divisible by 4 but will be divisible by 2

As
K is divisible by 8
we know k is even
so
(k^2 + 4k + 3)
will be odd
and odd no can never be divisible by 2
hence
(k + 2)(k^2 + 4k + 3) will not be divisible by 4
Sufficient


B
PS: I assumed k + 1/3 to be (k + 1)/3
since former does not make sense at all
it can not be an interger.

(k + 1)/3 is an odd integer
means
K is even
so K is divisible by 2
but we can not conclude about 4

so
K+2 also even and also divisible by 2
but we can not conclude about 4


K even so
(k^2 + 4k + 3) odd
hence we can not conclude whether exp is divisible by 4 or not
Insuff


Hope this helps

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by cramya » Thu Oct 23, 2008 1:24 pm
Explaining a little further


(k + 2)(k^2 + 4k + 3)

If u break down k^2+4k+3 u will get (k+1) (k+3)

So basically the question boils down to Is (k+1)(k+2)(k+3) is divisible by 4?

Stmt I) Given k is divisible by 8

It takes k+4 for the next number to be divisible by 4. Hence we can certainly conclude that (k+1)(k+2)(k+3) since its given k is divisible by 8


Eg: If k=8 it will be 9*10*11 - not divisible by 4

Stmt 2) k+1/3 is odd

3k+1/3 = odd
3k+1 = 3*odd

3*odd is odd(odd*odd = odd)

So 3k+1 is odd
In this 1 is odd so 3k has to be even(since odd+odd will give u even)
In 3k, 3 is odd for 3k to be even k has to be even(since odd*odd=odd whereas odd*even = even)

All we know is k is even

If k is 4 then (k+1)(k+2)(k+3) = 5*6*7 - Not divisible by 4
If k=2 (k+1)(k+2)(k+3) 3*4*5 - divisible by 4

INSUFFICEINT

Hence A)

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by cramya » Thu Oct 23, 2008 1:27 pm
ADDITIONS IN BOLD:

Explaining a little further


(k + 2)(k^2 + 4k + 3)

If u break down k^2+4k+3 u will get (k+1) (k+3)

So basically the question boils down to Is (k+1)(k+2)(k+3) divisible by 4?

Stmt I) Given k is divisible by 8

It takes k+4 for the next number to be divisible by 4. Hence we can certainly conclude that (k+1)(k+2)(k+3) is not divisible by 4 since its given k is divisible by 8

SUFFICIENT

Eg: If k=8 it will be 9*10*11 - not divisible by 4

Stmt 2) k+1/3 is odd

3k+1/3 = odd
3k+1 = 3*odd

3*odd is odd(odd*odd = odd)

So 3k+1 is odd
In this 1 is odd so 3k has to be even(since odd+odd will give u even)
In 3k, 3 is odd for 3k to be even k has to be even(since odd*odd=odd whereas odd*even = even)

All we know is k is even

If k is 4 then (k+1)(k+2)(k+3) = 5*6*7 - Not divisible by 4
If k=2 (k+1)(k+2)(k+3) 3*4*5 - divisible by 4

INSUFFICIENT

Hence A)

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Re: MGMAT DS

by yezz » Thu Oct 23, 2008 1:53 pm
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.

(k+2)(k+3)(k+1) ie: 3 consecutive intigers

in any 3 consecutive intigers ( devisiable by 4 if 2 of them r even or the midlle one is devisable by 4.

from one

if k is divisible by 8 , thus k+2 is even AND IMPOSSIBLE TO BE DEVISBLE BY 4

TRY ALL MULTIPLES OF 8X +2

from 2

k+1 = odd ie: k is even,, however we have no clue if (k+2 is a multiple of 4)

both

insuff.....A

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by rohangupta83 » Thu Oct 23, 2008 2:43 pm
good to see so many different approaches. imo A