How to prove that if $m = 10t+k $ and $67|t - 20k$ then 67|m? m, t, k are Natural numbers.
How can I prove that if $m = 10t+k $ and $67|t - 20k$ then 67|m ?

We have $k=m-10t$ and so $67 | t - 20(m - 10t) = 201t - m$. But $201 = 3 \times 67$, and so
$$67 | 67 \times 3t - m$$
And so $67 | m$.
This can also be seen using modular arithmetic: we have $m=10t+k$ and $t \equiv 20k \pmod {67}$, so $m \equiv 201k \pmod {67}$. But $201 = 3 \times 67$, so $m \equiv 0 \pmod {67}$, and so $67|m$.
