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

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

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Updated on: 26 Jun 2013, 09:33
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If m, n, and p are integers, is m + n odd?

(1) m = p^2 + 4p + 4
(2) n = p^2 + 2m + 1

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Chauahan Gaurav
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Originally posted by ichauhan.gaurav on 26 Jun 2013, 09:22.
Last edited by Bunuel on 26 Jun 2013, 09:33, edited 1 time in total.
RENAMED THE TOPIC.
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Re: If m, n, and p are integers, is m + n odd?  [#permalink]

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26 Jun 2013, 09:32
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If m, n, and p are integers, is m + n odd?

(1) m = p^2 + 4p + 4. Not sufficient.

(2) n = p^2 + 2m + 1. Not sufficient.

(1)+(2) Add (1) and (2) $$m+n=p^2 + 4p + 4 + p^2 + 2m + 1=2p^2+4p+2m+5=2(p^2+2p+m)+5=even+odd=odd$$. Sufficient.

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

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26 Jun 2013, 09:42
If m, n, and p are integers, is m + n odd?
(1) m = p^2 + 4p + 4
(2) n = p^2 + 2m + 1

Statement 1-
If p is Odd then m - odd + even + even = Odd
If p is Even then m - Even + even + even = Even
Insufficient (still value of n is missing)

Statement 2-
If p is Odd then n - odd + even + odd = Even
If p is Even then n - Even + even + odd = Odd
Insufficient (still value of M is missing)

Statement 1& 2-
If p is Odd then m = Odd
& n = Even
M+ n = Odd

If p is Even then m = Even
& n = Odd
M+ n = Odd

Sufficient

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

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06 Nov 2014, 17:44
Hi,

Why can't we factor M into (p+1)^2, which yields P = -1. Doesn't that lock in one of the variables to begin with? I realize that it won't give us a different answer but why isn't that a valid approach?
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Re: If m, n, and p are integers, is m + n odd?  [#permalink]

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07 Nov 2014, 04:31
russ9 wrote:
Hi,

Why can't we factor M into (p+1)^2, which yields P = -1. Doesn't that lock in one of the variables to begin with? I realize that it won't give us a different answer but why isn't that a valid approach?

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

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04 Sep 2015, 12:29
Bunuel,

If we try 0 for p then m+n is even. So shouldn't the answer be E

Could you clarify...
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Re: If m, n, and p are integers, is m + n odd?  [#permalink]

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05 Sep 2015, 04:11
aimtoteach wrote:
Bunuel,

If we try 0 for p then m+n is even. So shouldn't the answer be E

Could you clarify...

If p = 0, then:

m = p^2 + 4p + 4 = 4.
n = p^2 + 2m + 1 = 9.

m + n = 13 = odd.
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Re: If m, n, and p are integers, is m + n odd?  [#permalink]

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10 Nov 2016, 13:59
(1) m = p^2 +4p+4

p=0 --> m=4
p=1 --> m=9
p=2 --> m=16

NOT SUFFICIENT

(2) n=p^2 +2p+1

p=0 --> m=1
p=1 --> m=4
p=2 --> m=9

NOT SUFFICIENT

(1) + (2)
p=0 --> m+n=5
p=1 --> m+n=13

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

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10 Nov 2016, 16:42
If m, n, and p are integers, is m + n odd?

(1) m = p^2 + 4p + 4
(2) n = p^2 + 2m + 1

1. m = p^2 + 4p + 4

This doesn't tell us anything about N so its insuff.

However if you'd like to factor it you get (p+2)(p+2)=m
or p=m-2

2. n = p^2 + 2m + 1

Here you can plug in numbrs.
Depending on what you plug in n can be even or odd. Insuff

1+2
if you take the equation p=m-2 and plug it into the equation for statement 2 you get
m²-4m+4+2m+1=n
m²-2m+5=n

try plugging an even number for m (4)
16-8+5=13
e+o=odd

Now plug in an odd (3)

9-6=3+5=even

n and m are opposites so the sum is always odd
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Re: If m, n, and p are integers, is m + n odd?  [#permalink]

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27 Jan 2017, 19:12

1/ Rewrite m as (p+2)^2
2/ plug in m in n to express n solely in function of p: you get n=3p^2+8p+9
3/ Test for p=even if you get n+m=odd
4/ Test for p=odd if you get n+m=odd
5/ since 3 and 4 coincide (for p even and p odd we get n+m=odd), we can be sure using Statement 1 and 2 that m+n is odd.
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Re: If m, n, and p are integers, is m + n odd?  [#permalink]

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02 Mar 2018, 12:20
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Re: If m, n, and p are integers, is m + n odd? &nbs [#permalink] 02 Mar 2018, 12:20
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