Confirmation of a conjecture: $3\mid (8n)^5 - k$

Question

I have made a conjecture:

$$\forall n\in \mathbb{N} \setminus \{6r : r \in \mathbb{N}\}, \ 3\mid (8n)^5 - k$$

Such that $k =$ a constant containing all the digits of $(8n)^5$ except for the last $3$. The order of digits is unecessary.

e.g. Let $n = 1$.

$(8\times 1)^5 = 32768 \Rightarrow k_1 = 32\require{cancel}\cancel{768} \ \land \ k_2 = 23\cancel{768}$

$32768 - 32 = 32736 = 3\times 10912$

or

$32768 - 23 = 32745 = 3\times 10915$

$\therefore 3\mid (8n)^5 - k_1, k_2$

I do not know how to prove/disprove it apart from using computer power and trying to find a counter-example, however this has been tested for all $n \leqslant 10^6$ and I have not found one. Could somebody please help me?

Thank you in advance.

Answer 1

I obtain a contradiction for $n=7$. Then $(8n)^5=56^5=550731776$, and $56^5-550731=5\cdot23\cdot 4784183$.