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If m and n are integers, is n even? (1) m^2-n^2=2n-m (2) m

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If m and n are integers, is n even? (1) m^2-n^2=2n-m (2) m [#permalink] New post 01 Jul 2008, 07:51
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If m and n are integers, is n even?
(1) m^2-n^2=2n-m
(2) m is an even number.

A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
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Re: DS - Integers [#permalink] New post 01 Jul 2008, 08:07
Answer is (A):

(2) is clearly insufficient since it gives no information on n

(1) can be rewritten as (m-n)(m+n) = n - (m-n)

i.e. n = (m-n)(m+n+1)

Let's assume n can be odd. Then (m-n) and (m+n+1) are odd too

If n is odd and m-n is odd, then m is odd as well

If n is odd and m+n+1 is odd, then m+1 is odd, i.e. m is even

We are coming to a contradiction : n can not be odd and is thus even ==> sufficient
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Re: DS - Integers [#permalink] New post 01 Jul 2008, 08:17
when n is odd means (m-n) is odd too?
Can you explain this?
Are you assuming m to be even
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Re: DS - Integers [#permalink] New post 01 Jul 2008, 09:07
I agree with Oski till n = (m-n)(m+n+1)
then my argument is ... (m-n) and (m+n) are both either odd or even. And (m+n+1) is always going to be opposite of that.

hence n = (m-n)(m+n+1) = odd * even
or n = (m-n)(m+n+1) = even * odd
In any case, as one of its factors is even n is even.

and hence the Answer is (A)

Thanks Oski for the clue :)
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Re: DS - Integers [#permalink] New post 01 Jul 2008, 10:11
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Answer is A.

Stmt1: m^2-n^2=2n-m
or m^2 + m = 2n + n^2

Now, if m is odd, m^2 will be odd and hence m^2+m will be even.
Alternatively, if m is even, m^2 will be even and hence m^2+m will be even.

Thus, in either case, left hand side is even.

Now, 2n is always even. Thus, right hand side will be even only if n^2 is even. If n^2 is even, n will be even.
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Re: DS - Integers [#permalink] New post 01 Jul 2008, 10:34
This statement in not true

If n is odd and m-n is odd, then m is odd as well

if n=3, m-n=3, m=6 which is even.
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Re: DS - Integers [#permalink] New post 01 Jul 2008, 11:24
kapilnegi wrote:
If m and n are integers, is n even?
(1) m^2-n^2=2n-m
(2) m is an even number.

A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


I am getting A...though initially i thought the ans is C..but as i began to write out my work..it became clear to me A is sufficient

(m+n)(m-n)=n-m+n divide by (m+n)

(m-n)=n/(m+n) -1

now we know m and n are integers... therefore n/(m+n)=integer too..

m-n- n/(m+n)=-1

(m+n)(m) - n(m+n)-n=-(m+n)

m^2 +nm-nm-n^2-n=-m-n

m^2-n^2+m=0

m(m+1)=n^2
if m=even, n=even
m=odd, n=even
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Re: DS - Integers [#permalink] New post 01 Jul 2008, 11:26
nice work..but i dont think you need to assume anything here..
if you simplify you get m(m+1)=n^2..

m(m+1) is the product of 2 consecutive integers and thus will always be even..

Oski wrote:
Answer is (A):

(2) is clearly insufficient since it gives no information on n

(1) can be rewritten as (m-n)(m+n) = n - (m-n)

i.e. n = (m-n)(m+n+1)

Let's assume n can be odd. Then (m-n) and (m+n+1) are odd too

If n is odd and m-n is odd, then m is odd as well

If n is odd and m+n+1 is odd, then m+1 is odd, i.e. m is even

We are coming to a contradiction : n can not be odd and is thus even ==> sufficient
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Re: DS - Integers [#permalink] New post 01 Jul 2008, 11:28
Ans is A:

Here is my explanation:

m^2 - n^2 = 2n - m, can be written as
m^2 + m = 2n + n ^2
=> m (m + 1) = n (n+2)

m (m+1) need to be even becos, either m or (m+1) need to be even, as m is an integer.
So the product n (n + 2) need to be even. n & (n+2) can be the same either even or odd.
That leaves the option that n is even.

This is pretty gud question, First I went wrong on this, checked again and then came back with A.
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Re: DS - Integers [#permalink] New post 01 Jul 2008, 17:31
OA is A.

Thanks for the explainaton.

Cheers!
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Re: DS - Integers [#permalink] New post 01 Jul 2008, 17:41
Just a query.

What if m=n=0?

Statment 1 holds true for this as well.

Is '0' an even integer?
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Re: DS - Integers [#permalink] New post 01 Jul 2008, 20:20
kapilnegi wrote:
If m and n are integers, is n even?
(1) m^2-n^2=2n-m
(2) m is an even number


A. m^2 - n^2 = 2n - m
m^2 +m = n^2 + 2n ...................i
m (m + 1) = n (n + 2) .................. ii

i. m^2 + m is always even no matter m is even or odd.

if m is odd, m^2 is and m both are odd and sum of two odds is even.
if m is even, then there is no issue.

ii. m (m + 1) is also an even integer as m (m + 1) is a consecutive integers.
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Re: DS - Integers   [#permalink] 01 Jul 2008, 20:20
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