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Source: — Data Sufficiency |
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by neelgandham » Thu May 17, 2012 12:36 pm
If m, n, and p are integers, is m + n odd?
(1) m = p^2 + 4p + 4

No mention of n. So, Statement 1 is insufficient to answer the question.
(2) n = p^2 + 2m + 1
If p is an odd number and m is an odd number then n is an even number and m+n is an odd number
If p is an odd number and m is an even number then n is an even number and m+n is an even number.
We got two different answers. So, Statement 2 is Insufficient to answer the question.
1 + 2
Adding the left hand side and right hand side of both the statements, we get
m + n = 2*p^2 + 4p + 2m + 5 = 2*(p^2 + 2p + m) + 5 = 2X + 5( Where X is an Integer). So, m+n is always Odd.

IMO C
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by Anurag@Gurome » Thu May 17, 2012 6:44 pm
GmatKiss wrote: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 = p² + 4p + 4 = (p + 2)²
Now m can be either even or odd, depending on p's value and nothing is given about n; NOT sufficient.

(2) n = p² + 2m + 1
Now 2m will always be even, whether m is even or odd
p² + 1 can be either even or odd, which depends on p's value.
So, nothing is definite about m's and n's value; NOT sufficient.

Combining (1) and (2), m + n = p² + 4p + 4 + p² + 2m + 1 = 2p² + 4p + 2m + 5 = even + odd = odd
So, m + n is always odd; SUFFICIENT.

The correct answer is C.
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by minkathebest » Sat Sep 22, 2012 1:45 pm
Thank you!
Anurag@Gurome wrote:
GmatKiss wrote: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 = p² + 4p + 4 = (p + 2)²
Now m can be either even or odd, depending on p's value and nothing is given about n; NOT sufficient.

(2) n = p² + 2m + 1
Now 2m will always be even, whether m is even or odd
p² + 1 can be either even or odd, which depends on p's value.
So, nothing is definite about m's and n's value; NOT sufficient.

Combining (1) and (2), m + n = p² + 4p + 4 + p² + 2m + 1 = 2p² + 4p + 2m + 5 = even + odd = odd
So, m + n is always odd; SUFFICIENT.

The correct answer is C.
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