If m and n are integers, Is m odd?
1. n+m is odd.
2. n+m=n^2+5
Please help me on the above question, as i am not able to get the logic behind this!!
GMATPrep Question 28
This topic has expert replies
You can solve this by picking some numbers.
Choice 1:
n + m is odd
Sum of two integers is odd only if one of the numbers is odd and the other is even. So you can see that is the case n is even m has to be odd and in the case where n is odd m has to be even.
So 1 is not sufficient to answer if m is odd.
You can just pick numbers:
pick an odd value for n (say 3) and then m has to be even(say 4) to make the sum odd (7)
Similarly you can interchange values and see that if n is even (4) then m has to be odd (3) to make the sum odd (7)
Choice 2:
n + m = n^2 + 5
Pick an odd value for n in one example and an even value in the other.
See what happens to m.
Let's say n is 4(even) then 4 + m = 16 + 5, m = 17 (odd)
Now pick a odd value for n (say 5) then 5 + m = 25 + 5, m = 25 (odd)
So 2 is sufficient condition to say that m is odd.
The answer is B
Choice 1:
n + m is odd
Sum of two integers is odd only if one of the numbers is odd and the other is even. So you can see that is the case n is even m has to be odd and in the case where n is odd m has to be even.
So 1 is not sufficient to answer if m is odd.
You can just pick numbers:
pick an odd value for n (say 3) and then m has to be even(say 4) to make the sum odd (7)
Similarly you can interchange values and see that if n is even (4) then m has to be odd (3) to make the sum odd (7)
Choice 2:
n + m = n^2 + 5
Pick an odd value for n in one example and an even value in the other.
See what happens to m.
Let's say n is 4(even) then 4 + m = 16 + 5, m = 17 (odd)
Now pick a odd value for n (say 5) then 5 + m = 25 + 5, m = 25 (odd)
So 2 is sufficient condition to say that m is odd.
The answer is B
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(1) we know that to get an odd sum, one is odd and one is even, but we have no clue which is which: insufficient.gurudev wrote:If m and n are integers, Is m odd?
1. n+m is odd.
2. n+m=n^2+5
Please help me on the above question, as i am not able to get the logic behind this!!
(2) More complicated, pick numbers!
if n = 2, then we get:
2 + m = 4 + 5
m = 7. So "yes", m is odd.
if n = 3, then we get:
3 + m = 9 + 5
m = 11. So "yes", m is odd.
So, regardless of whether n is even or odd, m is always odd. (2) is sufficient alone, choose (B).
Using number properties instead of picking numbers:
n+m=n^2+5
m = n^2 - n + 5
m = n(n-1) + 5
well, since n is an integer, n and (n-1) are consecutive integers, which means one term is even and the other is odd. Therefore, regardless of the value of n, the product of n and (n-1) will always be even (since even*odd=even).
So:
m = even - 5
Since (even - odd) is always odd, m MUST be odd: sufficient.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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