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Algebra

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by heshamelaziry » Mon Nov 16, 2009 11:40 pm
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 ?
Last edited by heshamelaziry on Tue Nov 17, 2009 12:20 am, edited 1 time in total.
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Source: — Data Sufficiency |
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by italian7745 » Mon Nov 16, 2009 11:58 pm
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...
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by ershovici » Tue Nov 17, 2009 1:38 am
I also agree C looks to be correct.
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by viju9162 » Tue Nov 17, 2009 2:06 am
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
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by Gmat09_5ALL » Tue Nov 17, 2009 3:01 am
P^2 + P^2 = even
M+N = odd

Hence C
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by viju9162 » Tue Nov 17, 2009 4:18 am
Hi Gmat09_5ALL ,

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

Regards,
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by palvarez » Tue Nov 17, 2009 11:40 pm
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.
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by Abdulla » Tue Nov 17, 2009 11:52 pm
I agree with viju9162
I believe it's E .. what's the OA hesham?
Abdulla
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by linfongyu » Wed Nov 18, 2009 11:09 am
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.
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by chipbmk » Wed Nov 18, 2009 12:12 pm
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!
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by palvarez » Wed Nov 18, 2009 3:02 pm
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
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