jmc

We try finding the remainders when increasing powers of 5 are divided by 7. $$\begin{array}{rl}{5}^1÷7& \text{ leaves a remainder of }5.\\ 5^2÷7& \text{ leaves a remainder of }4.\\ 5^3÷7& \text{ leaves a remainder of }6.\\ 5^4÷7& \text{ leaves a remainder of }2.\\ 5^5÷7& \text{ leaves a remainder of }3.\\ 5^6÷7& \text{ leaves a remainder of }1.\\ 5^7÷7& \text{ leaves a remainder of }5.\\ 5^8÷7& \text{ leaves a remainder of }4.\end{array}$$ $$⋮$$ The remainders repeat after every 6 powers of 5. So we look for the remainder when 207 is divided by 6, which leaves a remainder of 3. We could use long division, but notice that 207 is a multiple of 3 (digits sum to 9, which is a multiple of 3) but is not a multiple of 2. That means 207 is not divisible by 6 and must be exactly 3 more than a multiple of 6. So the remainder for $5^{207}$ when divided by 7 is the same as the remainder when $5^3$ is divided by 7, which is $\boxed{6}$.