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If m and n are integers, is m odd?

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If m and n are integers, is m odd? [#permalink] New post 14 Jul 2010, 12:44
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If m and n are integers, is m odd?

(1) n + m is odd
(2) n + m = n^2 + 5
[Reveal] Spoiler: OA
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Re: m odd? [#permalink] New post 14 Jul 2010, 12:46
zisis wrote:
if m and n are integers, is m odd?

(1) n+m is odd
(2) n+m = n^2 + 5



1 is insuf
2 is insuf for me

let me explain:
m = n^2 - n + 5
thus, if n =0, m=1-0+5= 6 thus even
if n=1, m=1-1+5 = 5 thus odd...

what am i missing?
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Re: m odd? [#permalink] New post 14 Jul 2010, 12:55
Expert's post
(1) n,m could be odd,even ore even,odd. insufficient

(2) m = n^2 - n + 5 or m = odd|even - odd|even + odd = (odd - odd)|(even - even) + odd = even + odd = odd. sufficient.

zisis wrote:
if m and n are integers, is m odd?

let me explain:
m = n^2 - n + 5
thus, if n =0, m=1-0+5= 6 thus even
if n=1, m=1-1+5 = 5 thus odd...

what am i missing?


0^2 = 0.
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Re: m odd? [#permalink] New post 14 Jul 2010, 12:57
walker wrote:
(1) n,m could be odd,even ore even,odd. insufficient

(2) m = n^2 - n + 5 or m = odd|even - odd|even + odd = (odd - odd)|(even - even) + odd = even + odd = odd. sufficient.

zisis wrote:
if m and n are integers, is m odd?

let me explain:
m = n^2 - n + 5
thus, if n =0, m=1-0+5= 6 thus even
if n=1, m=1-1+5 = 5 thus odd...

what am i missing?


0^2 = 0.





aaaaaaaaaaaaaaaaaaaaaaaaaa
it s the other way round
2^0 = 1 and not 0^2 = 1
aaaaaaaaaa
going mental
thanks
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Re: m odd? [#permalink] New post 14 Jul 2010, 13:07
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If m and n are integers, is m odd?

(1) n + m is odd --> n+m=odd. The sum of two integers is odd only if one is odd and another is even, hence m may or may not be odd. Not sufficient.

(2) n + m = n^2 + 5 --> m-5=n^2-n --> m-5=n(n-1), either n or n-1 is even hence n(n-1)=even --> m-5=m-odd=even --> m=odd. Sufficient.

Answer: B.
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Re: If m and n are integers, is m odd? [#permalink] New post 16 Aug 2013, 05:22
m and n are integers, is m odd?
-Can't rephrase the question here, so you work with m = odd? It's a Yes/No DS question type.

Here's how I approached the problem.

Statement 1.
Given: m + n is odd.
Using even/odd number property rules:
even + odd = odd. OR, odd + even = odd. Since m can be either even or odd in this case, Statement 1 is insufficient, we don't have enough information.

Statement 2.
Given: m+n = n^2 + 5

Simplifying the equation:
m - 5 = n^2 - n
m - 5 = n(n - 1)
m = n(n-1) +5
m = Even + odd. Therefore, m is odd.

n(n-1) will always be even because it's a number times the number preceding it (think consecutive)...so one of those numbers has to be even (if n = 3 for example, you would have 3*(3-1) = 3*(2) = 6). You have an even*odd = even situation here.

5 is odd. So putting it all together, m = Even + Odd. You can answer with the given information that, yes, m is odd. Sufficient.

So the answer is B.
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Re: If m and n are integers, is m odd? [#permalink] New post 23 Nov 2013, 04:04
Statement 1, m and n could be both odd or one odd, one even. Insufficient.
Statement 2, when n is odd, n^2+5 is even, then m+n is even, m is odd; when n is even, n^2+5=odd, m+n is odd,
then m is odd. Sufficient.
So, answer is B
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Re: m odd? [#permalink] New post 28 Mar 2014, 00:11
Bunuel wrote:
If m and n are integers, is m odd?

(1) n + m is odd --> n+m=odd. The sum of two integers is odd only if one is odd and another is even, hence m may or may not be odd. Not sufficient.

(2) n + m = n^2 + 5 --> m-5=n^2-n --> m-5=n(n-1), either n or n-1 is even hence n(n-1)=even --> m-5=m-odd=even --> m=odd. Sufficient.
Answer: B.


Another method:
St1: n + m is odd, Answer is YES when n is even but answer is NO when n is odd.
Insufficient!

St2: n + m = n^2 + 5 --> n^2 - n + (5-m) = 0 -- > is a quadratic equation with sum of roots = 1 and product of roots = (5-m). Clearly the the roots are consecutive integers with least among them as negative (ex. 4 & -3, 6 & -5...) That means one of the root is even. So, product of roots (5-m) is also even. For (5-m) to be even, m must be odd.
Sufficient!
Answer: B
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Re: If m and n are integers, is m odd? [#permalink] New post 28 Mar 2014, 07:22
Satatment 1:Clearly insuff.
Even+odd=odd.So m could be even or odd.
S2:sufficient.
The eq can be rearranged as
N(n-1)=m-5
Now n(n-1) is a product of 2 consecutive integers so it'll definitely be even=>m-5=even
So m=odd because only odd-odd=even
Ans option B

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Re: If m and n are integers, is m odd?   [#permalink] 28 Mar 2014, 07:22
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