GMATPrep Question 28

[gurudev]

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!!

[bourne159]

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

[Stuart@KaplanGMAT]

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!!

(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.

(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.

[gurudev]

Yeah Got it now!!!!

Thanks a ton..