While working with a larger proof, I wanted to prove the following result which I hope is true.
The question.
Let $(e_k)$ be a sequence which takes it values in $\{1,2\}$.
Let $\gamma,\delta\in\{0,\ldots,7^n\}$ such that $(\gamma,\delta)\notin\{(0,0),(0,7^n),(7^n,0),(7^n,7^n)\}$.
Assuming $$K=\sum_{k=0}^n \frac{\gamma e_k+\delta}{7^k}\in\mathbb Z,$$ does this imply $$\frac{\gamma e_n+\delta}{7^n}\in\mathbb Z\qquad ?$$
What I tried.
I have showed that without loss of generality, we can assume $\gcd(7,\gamma)=\gcd(7,\delta)=1$.
We can also assume that $7\mid \gamma e_n+\delta$ or it is straightforward.
Regarding Dietrich Burde's comment, if $\gamma=0$, then $\delta\ne 7^n$, so if we observe the $7$-adic development of $K$, it as a digit $<1$, which is absurd.
Any hints or leads would be much appreciated.
I found a counter-example:
$$(7^2-1)+8+\frac{2(7^2-1)+8}7+\frac{(7^2-1)+8}{7^2}\in\mathbb Z$$
but
$$\frac{(7^2-1)+8}{7^2}=1+\frac{1}{7}\notin\mathbb Z.$$