How would I prove that if 11n-5 is odd, then n is an even number using only a direct proof?

Question

I've been able to prove this statement through contraposition and contradiction but I'm struggling to prove it through a direct proof. It seems I always get it in the form where 11n=2(k+3).

Answer 1

Since the right-hand side has $2$ as a factor, and $2$ and $11$ are coprime, the left-hand side also has $2$ as a factor. This relies on the fundamental theorem of arithmetic, where every positive integer can be represented as a unique product of prime numbers.

Answer 2

Lets assume $11n-5$ is odd then there exists integer $k$ such that

$11n-5=2k+1 \implies 11n=2k+1+5 = 2k+6 = 2(k+3)$.

Now as $k$ is an integer we get $k+3$ is an integer. We can set $m = k+3$ then we get $11n = 2m$ and $2m$ is an even number which implies $11n$ is an even number. As we have found $11n$ to be an even number this implies that $n$ must be even since the product of two odd numbers will be odd.

Answer 3

$11n - 5 \equiv 1\ (\textrm{mod}\ 2)$

$11n \equiv 0\ (\textrm{mod}\ 2)$

Since $11 \equiv 1\ (\textrm{mod}\ 2)$,

$11n \equiv 1 * n \equiv n \equiv 0\ (\textrm{mod}\ 2)$.