If m, n, and p are integers, is m + n odd?
(1) m = p2 + 4p + 4
(2) n = p2 + 2m + 1
OAC
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If m, n, and p are integers, is m + n odd?
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GMATGuruNY
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Statement 1:rsarashi wrote:If m, n, and p are integers, is m + n odd?
(1) m = p² + 4p + 4
(2) n = p² + 2m + 1
No information about n.
INSUFFICIENT.
Statement 2:
Case 1: p=0, m=0, n=1
Here, m+n = 0+1 = 1, with the result that the answer to the question stem is YES.
Case 2: p=1, m=0, n=2
Here, m+n = 0+2 = 2, with the result that the answer to the question stem is NO.
INSUFFICIENT.
Statements combined:
m = (p+2)².
n = p² + 2m + 1.
Case 1: p = 0
In this case:
m = (0+2)² = 4.
n = 0² + 2*4 + 1 = 9.
m+n = 4+9 = 13.
Here, the answer to the question stem is YES.
Case 2: p = 1
In this case:
m = (1+2)² = 9.
n = 1² + 2*9 + 1 = 20.
m+n = 9+20 = 29.
Here, the answer to the question stem remains YES.
Cases 1 and 2 indicate that -- whether p is EVEN or ODD -- the answer to the question stem will always be YES.
SUFFICIENT.
The correct answer is C.
Last edited by GMATGuruNY on Mon May 29, 2017 5:25 am, edited 1 time in total.
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Brent@GMATPrepNow
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I assume that statement 2 should be as follows...
Target question: Is m + n odd?
Statement 1: m = p² + 4p + 4
No information about n.
So, there's no way to answer the target question with certainty.
Statement 1 is NOT SUFFICIENT
Statement 2: n = p² + 2m + 1
No information about n.
So, there's no way to answer the target question with certainty.
Statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
Since both m and n both rely on the value of p, and since p is EITHER an odd integer or an EVEN integer. Let's examine each possible case.
Case a: p is ODD.
m = p² + 4p + 4 = (p +2 )² = (ODD +2 )² = (ODD )² = ODD
n = p² + 2p + 4 = (p +1 )² = (ODD +1 )² = (EVEN)² = EVEN
So, if p is odd, m+n = ODD + EVEN = ODD
Case b: p is EVEN.
m = p² + 4p + 4 = (p +2 )² = (EVEN+2 )² = (EVEN)² = EVEN
n = p² + 2p + 4 = (p +1 )² = (EVEN +1 )² = (ODD )² = ODD
So, if p is odd, m+n = EVEN + ODD = ODD
So, regardless of whether p is even or odd, m+n will ALWAYS be ODD
Since we can answer the target question with certainty, the combined statements are SUFFICIENT
Answer: C
Cheers,
Brent
rsarashi wrote:If m, n, and p are integers, is m + n odd?
(1) m = p² + 4p + 4
(2) n = p² + 2p + 1
OAC
Target question: Is m + n odd?
Statement 1: m = p² + 4p + 4
No information about n.
So, there's no way to answer the target question with certainty.
Statement 1 is NOT SUFFICIENT
Statement 2: n = p² + 2m + 1
No information about n.
So, there's no way to answer the target question with certainty.
Statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
Since both m and n both rely on the value of p, and since p is EITHER an odd integer or an EVEN integer. Let's examine each possible case.
Case a: p is ODD.
m = p² + 4p + 4 = (p +2 )² = (ODD +2 )² = (ODD )² = ODD
n = p² + 2p + 4 = (p +1 )² = (ODD +1 )² = (EVEN)² = EVEN
So, if p is odd, m+n = ODD + EVEN = ODD
Case b: p is EVEN.
m = p² + 4p + 4 = (p +2 )² = (EVEN+2 )² = (EVEN)² = EVEN
n = p² + 2p + 4 = (p +1 )² = (EVEN +1 )² = (ODD )² = ODD
So, if p is odd, m+n = EVEN + ODD = ODD
So, regardless of whether p is even or odd, m+n will ALWAYS be ODD
Since we can answer the target question with certainty, the combined statements are SUFFICIENT
Answer: C
Cheers,
Brent
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Jay@ManhattanReview
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(m+n) would be odd if one of m and n is odd and the other is even.rsarashi wrote:If m, n, and p are integers, is m + n odd?
(1) m = p^2 + 4p + 4
(2) n = p^2 + 2m + 1
OAC
Statement 1: m = p^2 + 4p + 4
=> m+n = (p^2 + 4p + 4) + n = (p^2 + n) + EVEN + EVEN; since (p^2 + n) can be even or odd, we cannot determine whether m+n is odd.
Statement 2: n = p^2 + 2m + 1
=> m+n = (p^2 + 2m + 1) + m = (p^2 + 3m) + ODD; since (p^2 + 3m) can be even or odd, we cannot determine whether m+n is odd.
Statement 1 & 2:
Adding Statement 1 and 2, we get
m+n = (p^2 + 4p + 4) + (p^2 + 2m + 1)
=> m+n = 2p^2 + 4p + 2m + 5 = EVEN + EVEN + EVEN + ODD = ODD.
Thus, m+n is odd. Sufficient.
The correct answer: C
Hope this helps!
-Jay
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masoom j negi
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m + n will be odd if one of them is odd.
Statement 1. m = p2 + 4p + 4 = (p + 2)2
So, m is even if p is even and m is odd if p is odd.
But we can't say anything about m+n. Hence, Insufficient.
Statement 2. n = p2 + 2m + 1.
2m is even because it is a multiple of 2.
P2 + 1 will be even if p is odd and p2 +1 will be odd if p is even.
So, n is even if p is odd and n is odd if p is even.
But we can't say anything about m +n. Hence, Insufficient.
Statement 1 & 2 together. Using the results of statement 1 & 2, we can say that
If p is even: m is even and n is odd.
If p is odd: m is odd and n is even.
Hence, m + n will always be odd. Hence, Sufficient.
Statement 1. m = p2 + 4p + 4 = (p + 2)2
So, m is even if p is even and m is odd if p is odd.
But we can't say anything about m+n. Hence, Insufficient.
Statement 2. n = p2 + 2m + 1.
2m is even because it is a multiple of 2.
P2 + 1 will be even if p is odd and p2 +1 will be odd if p is even.
So, n is even if p is odd and n is odd if p is even.
But we can't say anything about m +n. Hence, Insufficient.
Statement 1 & 2 together. Using the results of statement 1 & 2, we can say that
If p is even: m is even and n is odd.
If p is odd: m is odd and n is even.
Hence, m + n will always be odd. Hence, Sufficient.
