What is the minimum number of groups of three coins you can flip in order to flip every coin so that it is heads-down? 
Question: Suppose that there are seven coins arranged in a circle. Every coin is heads-up. Your goal is to flip every coin so that it is heads-up. You may, however, only flip groups of three adjacent coins. You may flip any such group of three coins, and you map flip as many such groups as you choose, one group at a time. What is the minimum number of groups of three coins you can flip in order to flip every coin so that it is heads-down?

My attempt:  
Let $x$ be the number of groups to be flipped.
So we can interpret the question as 
$$3x≡0 \,\,\,mod \,\,\,7.$$
One possible solution is $x=7$. So it requires at most 7 groups.
On the other hand, clearly it requires at least $3$ groups. So minimum is between $3$ and $7$.
Then I am stuck here. 
Any hint is appreciated.

Label the coins $a$ to $g$, and the groups of three coins $A$ to $G$ where the label corresponds to the middle coin.
Flipping the same group twice is equivalent to doing nothing, so we flip each group either $0$ or $1$ times.
Each coin must be flipped an odd number of times. Suppose wlog that coin $a$ is flipped only once. What can we say about the number of times that $g$ and $b$ are flipped?
