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# If m and n are integers, is m odd?

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Manager
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If m and n are integers, is m odd? [#permalink]

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14 Jul 2010, 13:44
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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
Manager
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14 Jul 2010, 13:46
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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14 Jul 2010, 13:55
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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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14 Jul 2010, 13:57
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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14 Jul 2010, 14:07
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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.

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Re: If m and n are integers, is m odd? [#permalink]

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16 Aug 2013, 06:22
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.

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Re: If m and n are integers, is m odd? [#permalink]

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23 Nov 2013, 05:04
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.
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28 Mar 2014, 01:11
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.

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!
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Re: If m and n are integers, is m odd? [#permalink]

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28 Mar 2014, 08:22
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]

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12 May 2015, 02:17
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Re: If m and n are integers, is m odd? [#permalink]

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14 May 2015, 17:02
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Expert's post
Hi All,

Even though this is an old post (and most of the original posters are probably long gone), this question serves as a nice example of some of the Number Properties that you're likely to face on Test Day. If you can spot the NPs involved, then you can move relatively quickly through the work, but even if you don't recognize the NPs, you can still TEST VALUES to prove that patterns exist. In that way, you can do some quick work and avoid "staring at the screen" and hoping that something comes to you.

We're told that M and N are INTEGERS. We're asked if M is ODD. This is a YES/NO question.

Fact 1: N + M is ODD

IF....
N = 1
M = 0
The answer to the question is NO

IF....
N = 0
M = 1
The answer to the question is YES
Fact 1 is INSUFFICIENT

Fact 2: N + M = N^2 + 5

Here, the value of M depends on the value of N. Let's start with something simple and see if a pattern emerges.

IF....
N = 0
M = 5
The answer to the question is YES

IF....
N = 1
M = 5
The answer to the question is YES

IF....
N = 2
M = 7
The answer to the question is YES

IF....
N = 3
M = 11
The answer to the question is YES

It certainly looks like the answer to the question is ALWAYS YES. We can quickly TEST some negative values for N and see what happens....

IF....
N = -1
M = 7
The answer to the question is YES

IF....
N = -2
M = 11
The answer to the question is YES
Fact 2 is SUFFICIENT

[Reveal] Spoiler:
B

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Re: If m and n are integers, is m odd? [#permalink]

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15 May 2015, 00:48
very tricky, I missed this question.

not hard but enough to kill us
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Re: If m and n are integers, is m odd? [#permalink]

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02 Jun 2016, 05:29
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Re: If m and n are integers, is m odd? [#permalink]

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02 Jun 2016, 10:25
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zisis wrote:
If m and n are integers, is m odd?

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

1) n + m is odd
It can be odd+even or even +odd. m can be even or odd. Not sufficient.

(2) n + m = n^2 + 5

Let's suppose n is odd, n^2 will also be odd. Hence to maintain the equality m must be odd.

Let's suppose n is even, n^2 will be even. And to maintain the equality m must be odd.

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Re: If m and n are integers, is m odd? [#permalink]

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11 Oct 2016, 05:07
zisis wrote:
If m and n are integers, is m odd?

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

A) m + n = odd
Two possibilities :-
1- m is odd and n is even : 1 + 2 = 3
2- m is even and n is odd : 2 + 1 = 3
Two possibilities, :. Insufficient

B) m+n = n^2 + 5

Solve

m + n - n^2 = 5
m + n(1-n) = 5
n(1-n) {can say one is odd other is even like consecutive numbers}

:. m + n(1-n) [even] = 5 [odd]

This is only possible when m is odd
as Odd + Even = Odd

Hence (B)
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Re: If m and n are integers, is m odd? [#permalink]

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21 Nov 2016, 11:15
Great Question
Here we are given two integers m,n
And we are asked if m is odd or not
Statement 1
m+n=odd
hmm
This Statement just tells us that m and n are of opposite even/odd nature
Not sufficient
Statement 2
m=n(n-1)+5
here n(n-1) is a product of two consecutive integers. hence it must be even
so n=even+5 => even+odd => odd
hence sufficient
hence B
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Re: If m and n are integers, is m odd? [#permalink]

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21 Nov 2016, 11:28
zisis wrote:
If m and n are integers, is m odd?

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

FROM STATEMENT - I ( INSUFFICIENT )

If n + m = Odd

Either n = Odd/Even and m = Odd/Even because -

1. Even + Odd = Odd
2. Odd + Even = Odd

FROM STATEMENT - II ( INSUFFICIENT )

n - n^2 = 5 - m

Or, n ( n - 1 ) = 5 - m

Here, LHS must be Even, because product of 2 consecutive integers is always even ( ie, one of the 2 consecutive numbers must be even )

So, 5 - m = Evem

Or, Odd - m = Even

Hence, m = Odd

Thus, Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked, answer will certainly be (B)...

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Re: If m and n are integers, is m odd?   [#permalink] 21 Nov 2016, 11:28
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