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!!
(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.
