#9 Hua Luo Geng.

We note that $a = 2^{2017}\times 3 \times 5^{2017}$ has the prime factors 2, 3 and 5 and that:

$\begin{aligned} b & \equiv 2 \times 10^{2017} + 1 \equiv 1 \text{ (mod 2)} \implies \color{#D61F06} 2 \not \mid b \end{aligned}$

$\begin{aligned} b & \equiv 2 \times 10^{\color{#3D99F6}2017 \text{ mod }\phi (3)} + 1 \text{ (mod 3)} & \small \color{#3D99F6} \text{Since }\gcd(3,10)=1 \text{, Euler's theorem applies.} \\ & \equiv 2 \times 10^{\color{#3D99F6}2017 \text{ mod }2} + 1 \text{ (mod 3)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (3) = 2 \\ & \equiv 2 \times 10^{\color{#3D99F6}1} + 1 \text{ (mod 3)} \\ & \equiv 21 \equiv 0 \text{ (mod 3)} \implies \color{#3D99F6} 3 \mid b \end{aligned}$

$\begin{aligned} b & \equiv 2 \times 10^{2017} + 1 \equiv 1 \text{ (mod 5)} \implies \color{#D61F06} 5 \not \mid b \end{aligned}$

Therefore, 3 is the only common factor of $a$ and $b$ and hence $\gcd(a,b)=3$ .

Using the Euclidean algorithm

$\begin{aligned} \gcd(3\cdot 10^{2017},2\cdot 10^{2017}+1)&=\gcd(3\cdot 10^{2017}-(2\cdot 10^{2017}+1),2\cdot 10^{2017}+1)\\ &=\gcd(10^{2017}-1,2\cdot 10^{2017}+1)\\ &=\gcd(10^{2017}-1,2\cdot 10^{2017}+1-2(10^{2017}-1))\\ &=\gcd(10^{2017}-1,3) \end{aligned}$

Which is $1$ or $3$ , depending on whether $10^{2017}-1$ is divisible by $3$ or not

$10^{2017}-1\equiv(3\cdot 3+1)^{2017}-1\equiv 1^{2017}-1\equiv 0\mod 3$

Hence

$\gcd(3\cdot 10^{2017},2\cdot 10^{2017}+1)=\boxed{3}$