Remember Fermat?

Applying Fermat's little theorem : $2^6 \equiv 1 \text{ mod 7} \Rightarrow 2^{20} = (2^6)^{3}\cdot 2^2 \equiv 4 \text{ mod 7}$ $3^6 \equiv 1 \text{ mod 7} \Rightarrow 3^{30} = (3^6)^{5} \equiv 1 \text{ mod 7}$ $4^6 \equiv 1 \text{ mod 7} \Rightarrow 4^{40} = (4^6)^{6}\cdot 4^4 \equiv 4 \text{ mod 7}$ $5^6 \equiv 1 \text{ mod 7} \Rightarrow 5^{50} = (5^6)^{8} \cdot 5^2 \equiv 4 \text{ mod 7}$ $6^6 \equiv 1 \text {mod 7} \Rightarrow 6^{60} = (6^6)^{10} \equiv 1 \text{ mod 7}.$ Therefore, $2^{20} + 3^{30} + 4^{40} + 5^{50} +6^{60} \equiv 4 +1 + 4 + 4 + 1 = 14 \equiv 0 \text{ mod 7}$

Since 2, 3, 4, 5, and 6 are coprime to 7, we can apply Euler's theorem . We note that Euler's totient function $\phi(7) = 6$ . Then, we have:

$\begin{aligned} x & \equiv \left(2^{20} + 3^{30} + 4^{40} + 5^{50} + 6^{60} \right) \text{(mod 7)} \\ & \equiv \left(2^{\color{#D61F06}{3\times 6}+2} + 3^{\color{#D61F06}{5\times 6}} + 4^{\color{#D61F06}{6\times 6}+4} + 5^{\color{#D61F06}{8\times 6}+2} + 6^{\color{#D61F06}{10\times 6}} \right) \text{(mod 7)} \\ & \equiv \left(\color{#D61F06}{\left(2^{\phi(7)} \right)^3}2^2 + \color{#D61F06}{\left(3^{\phi(7)} \right)^5} + \color{#D61F06}{\left(4^{\phi(7)} \right)^6}4^4 + \color{#D61F06}{\left(5^{\phi(7)} \right)^8}5^2 + \color{#D61F06}{\left(6^{\phi(7)} \right)^{10}} \right) \text{(mod 7)} \\ & \equiv \left(\color{#D61F06}{1}\cdot 2^2 + \color{#D61F06}{1} + \color{#D61F06}{1}\cdot 4^4 + \color{#D61F06}{1}\cdot 5^2 + \color{#D61F06}{1} \right) \text{(mod 7)} \\ & \equiv \left(2^2 + 1 + 4^4 + 5^2 + 1 \right) \text{(mod 7)} \\ & \equiv \left(4 + 1 + 2^{6+2} + 25 + 1 \right) \text{(mod 7)} \\ & \equiv \left(4 + 1 + 4 + 4 + 1 \right) \text{(mod 7)} \\ & \equiv 14 \text{(mod 7)} \\ & \equiv \boxed{0} \text{(mod 7)} \end{aligned}$