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Mishari
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Mishari
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.

So the answere is D.

Javed.

Cheers!
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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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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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javed
Mishari
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.

So the answere is D.

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! :)
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Good Job everybody

OA: A



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