IMO C...

m + n can be odd only when one of them is even and one of them is odd..

Now

1) m = (p+2)^2

so let p=0 ...m=4
let p = 1 ...m=9...

we do not know anything about n ....

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

here we have to assume both p and m ...

let m=1 odd ...p =1 ....n = 4... so m + n is odd....
let m =2 ....p=1 .... n= 6...so m + n is even...

so this is insufficient..

Now if we combine them...

m = (p + 2)^2
n= P^2 + 2 ( P^2 + 4P + 4) + 1
= 3P^2 + 8P + 9

if p = 1 .... m = 9.... n = 20... m + n is odd...
p=0 .... m = 4...... n = 9.... m + n is odd...

I also agree C looks to be correct.

I believe answer is E.

My reasoning is mentioned below. Please correct me if I am wrong.

Question - Is m+n = odd. For m+n to be odd, both m&n should not be even. And also, m&n should not be odd.

From A, m = p^2 + 4p + 4

m = p^2 + ( E) .. because 4p is even ( any number multiplied by 4 is even, hence 4p is even)and 4 is even ( E+E=E).

Therefore, m can be:

m = O+E
m= odd

because p^2 can be either even or odd . 3^2 = 9 or 2^2 = 4.

Not sufficient

From B, n = p^2 + 2m + 1

2m+1 = E+O = 0 ... 2m is always even.

Therfore, n can be:

n = p^2 + O

p^2 can be either even or odd .. therefore n can be even or odd.

From C, combining both:

m and n can either be both odd, which results to even integer. Or it can be odd as one of them can be odd integer.

Hence answer is E.

Regards,
Viju

P^2 + P^2 = even
M+N = odd

Hence C

Hi Gmat09_5ALL ,

Thanks. got it! I should have realized it.. went through long procedure to solve .

Regards,
Viju

heshamelaziry wrote:If m, n, and p are integers, is m + n odd?

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

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

Not sure how go about this one ?

Rephrase it

m+n = 1 (mod 2)


1. m = p^2 + 4p +4
m = p^2 (mod 2)

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


m +n = 2p^2 + 1 = 1 (mod 2).

Get rid of the junk in "rephrasing phase" itself. Carrying it over creates all troubles.

I agree with viju9162
I believe it's E .. what's the OA hesham?

Stem 1: m = p^2 + 4p + 4
m = (p+2)(p+2)
m is even if p is even, or odd if p is odd
No mention of n, eliminate A and D.

Stem 2: n = p^2 + 2m + 1
if p is odd, n = odd + even + odd = even
if p is even, n = even + even + odd = odd
Insufficient, eliminate C

Combine:
N depends on the p. No new information added regarding P. Choose E.

palvarez wrote:

heshamelaziry wrote:If m, n, and p are integers, is m + n odd?

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

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

Not sure how go about this one ?

Rephrase it

m+n = 1 (mod 2)


1. m = p^2 + 4p +4
m = p^2 (mod 2)

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


m +n = 2p^2 + 1 = 1 (mod 2).

Get rid of the junk in "rephrasing phase" itself. Carrying it over creates all troubles.

Can you explain what you mean when you write "(mod 2)"

Thanks!

chipbmk wrote:

palvarez wrote:

heshamelaziry wrote:If m, n, and p are integers, is m + n odd?

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

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

Not sure how go about this one ?

Rephrase it

m+n = 1 (mod 2)


1. m = p^2 + 4p +4
m = p^2 (mod 2)

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


m +n = 2p^2 + 1 = 1 (mod 2).

Get rid of the junk in "rephrasing phase" itself. Carrying it over creates all troubles.

Can you explain what you mean when you write "(mod 2)"

Thanks!

https://en.wikipedia.org/wiki/Modular_arithmetic