Prove or disprove: Removing $4$ distinct $K_3$s from $K_6$ leaves no other $K_3$.
I tried this and I believe it is true and to prove it I noted that when we remove $4K_3$s we are left with $3$ edges which are not incident to the same vertices. I don't know how to say this more precisely.. I think that each of the vertices is involved in $2K_3$s and at the end they have degree $1$. I just need to formulate this nicely. Any help is appreciated!
The reason why all vertices must have degree $1$ at the end is a parity argument.
At the beginning, all vertices have degree $5$, which is odd. Every time you take out a $K_3$, you decrease the degrees of three vertices by $2$, so all degrees remain odd. So after you've taken out four $K_3$'s, the degrees of the vertices are $6$ odd numbers that sum up to $6$ (because there are $3$ edges remaining). This can only happen by taking $1+1+1+1+1+1 = 6$, so all degrees are $1$ at the end.
(Similarly, we can show that in any graph that can be decomposed into $K_3$'s, all vertex degrees must be even. So it might be possible with $K_5$ and $K_7$ (turns out that it is possible for $K_7$, but not for $K_5$) but definitely isn't possible for $K_6$.)