Given $\,27\mid n = 54 \cdot2^2k+27k^2-27k,\,$ how to deduce $\,54\mid n $?

Question

My question relates to this problem:

Prove by induction that 54 divides $2^{2k+1}-9k^2+3k-2$.

My solving so far gives this answer: (after all calculations)

$2^{2(k+1)+1}-9(k+1)^2+3(k+1)-2= 54 \cdot2^2k+27k^2-27k$

It is obvious that $27=\frac{1}{2}54$ divides this expression, but how do I figure it out if 54 divides it too? The end result is correct (checked!)

Answer 1

HINT

Note that the first and last terms are even and hence are divisible by 2. The middle two are really $$ 3k-9k^2 = 3k(1-3k), $$ and the factors always have different parity, hence one of them is always even as well.

Answer 2

$27k^2-27k = 27k(k-1)$.

Exactly one of $k$ and $k-1$ is even, so...