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# k and m are positive integers such that if k is divided by m

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k and m are positive integers such that if k is divided by m [#permalink]

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17 Apr 2007, 15:04
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k and m are positive integers such that if k is divided by m the remainder is an odd integer.

Is k an even integer ?

(1) When k is divided by m, the quotient is even
(2) m is odd
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17 Apr 2007, 17:44

from stem: k=bm+o where o is odd.
st1 claims b is even, hence k is odd and answer to stem is NO. but sufficient.
st2 m is odd, b can be either even or odd, hence bm can be either even or odd hence k can be either odd or even - hence insufficient.
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17 Apr 2007, 23:45
Statement A is sufficient but B is not, SO A is winner
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Re: DS - Even and Odd [#permalink]

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18 Apr 2007, 00:07
Mishari wrote:
k and m are positive integers such that if k is divided by m the remainder is an odd integer.

Is k an even integer ?

(1) When k is divided by m, the quotient is even
(2) m is odd

Let o represent odd nos and e represent e events. now form the question stem we have k=nm+o. Where n represents all the odd and even nos.

From statement 1 we have k=em+o. Therefore em will be an even number and o will an odd no and hence k will be odd. So sufficient.

From statement 2 we have k=no+o. Now n can be even or odd. if n is even then k is odd and if n is odd then also k is odd. So again sufficient.

Javed.

Cheers!
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18 Apr 2007, 01:27
k = mq1 + R where R is odd.

St1:
If q1 is even, then q1m is even. Then q1m + R = even + odd = odd. So k is odd. Sufficient.

St:2
Insufficient as q1 can change the even/odd state of mq1 and hence k.

Ans A
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18 Apr 2007, 13:07
I am going with A..

K=Mx+R

we know R is odd..so if MX is even then even+odd=odd

1) MX is even...sufficient K is odd

2) M is odd..well depending on the value of x ..MX can be even or ODD..we dont know..Insuff

A it is
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Re: DS - Even and Odd [#permalink]

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18 Apr 2007, 15:18
javed wrote:
Mishari wrote:
k and m are positive integers such that if k is divided by m the remainder is an odd integer.

Is k an even integer ?

(1) When k is divided by m, the quotient is even
(2) m is odd

Let o represent odd nos and e represent e events. now form the question stem we have k=nm+o. Where n represents all the odd and even nos.

From statement 1 we have k=em+o. Therefore em will be an even number and o will an odd no and hence k will be odd. So sufficient.

From statement 2 we have k=no+o. Now n can be even or odd. if n is even then k is odd and if n is odd then also k is odd. So again sufficient.

Javed.

Cheers!

Javed, A should be the answer.

Try picking numbers for stmt 2. For example,

K= 8 M=7
8/7=1R1

Now change K to 15

15/7= 2R1

K can be either even or odd.

Hope this helps!
Director
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20 Apr 2007, 08:19
Good Job everybody

OA: A
20 Apr 2007, 08:19
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