Is the integer n odd? (1) n^2 - 2n is not a multiple of 4 (2) n is a

1. \(n^2-2n\) is not a multiple of 4.

follows \(n(n-2)\)is not a multiple of 4

find out in which points the equation yields zero, zero is a multiple of 4 as well as a multiple of any number except zero itself.
value a=0 and value b=2 thus n must not be any of those two values because otherwise the expression would result in a multiple of 4
-n can be both positive or negative, the only restriction is that n is an integer- since o and 2 are out of the pool 4 would be the first positive even number applicable to n. Every other positive even number would cause the expression to be a multiple of 4. We can safely say that n is for sure not an even number. before submitting the answer let's quickly check how the expression behaves with negatives.

if n=-2 the greatest negative even integer plugged \(n^2-2n\) results in a multiple of 4 then for sure n is odd.

Sufficient.


2. n is a multiple of 3.
Non sufficient, n could be zero.