We use the following facts.
$c_{12}=59$.
(Suppose that $c_{12}=53$.
If $c_9=59$, then either $c_5$ or $c_8$ has to be $53$, which is a contradiction. So, $c_9$ is of the form $\circ 3$.
Then, $c_3,c_4,c_5,c_7,c_8,c_9,c_{12}$ are of the form $\circ 3$. From fact 1, this is impossible.)
$c_9=A9$.
(Suppose that $c_9=53$. Then, since $c_{16}$ is of the form $\circ 9$, $c_{21}$ is of the form $\circ 3$. So, $c_3,c_4,c_5,c_7,c_8,c_9,c_{21}$ are of the form $\circ 3$. From fact 1, this is impossible.)
$c_{21}=B3$.
(If $c_{16}=53$, then $c_{21}$ is of the form $\circ 3$.
If $c_{16}$ is of the form $\circ 9$, then $c_{21}$ is of the form $\circ 3$.)
$c_3=C3$.
$c_{11}=D9$.
(If $c_5=29$, then since $c_7,c_8$ are of the form $\circ 3$, $c_{11}$ is of the form $\circ 9$.
If $c_8=29$, then since $c_7$ is of the form $\circ 9$, $c_{11}$ is of the form $\circ 9$.)
$C\not=1$.
(Suppose that $C=1$. Then, $B\not=1$ because of $c_{21}=B3$. Also, we have $c_{20}=B1,c_{19}=E1,c_{18}=41$.
If $c_5=29$, then we have $c_8=A3,c_7=D3$, so $B=7$ because $c_{20}=B1,c_{21}=B3,c_3=13,c_{18}=41$. Also, $A=8$ because $c_8=A3,c_9=A9,c_3=13,c_4=23,c_{12}=59,c_{21}=73$. However, there is no such $D$ because $c_7=D3,c_{11}=D9,c_3=13,c_4=23,c_{12}=59,c_{21}=73,c_8=83$. So, $c_8=29$.
We have $c_5=A3$ because $c_4=23,c_8=29,c_9=A9$. Also, $c_7=19$ because $c_8=29,c_{11}=D9,c_3=C3$. $$
\begin{array}{c|c|c|c|c}
& & 13 & 23 & A3\\\hline
& \times & 19 & 29 & A9\\\hline
& \times & D9 & \times & 59 \\\hline
& & &\times & \\\hline
& 41 & E1 & B1 & B3\end{array}
$$ So, $B=7$ because $c_{20}=B1,c_{21}=B3,c_3=13,c_4=23,c_{18}=41$. Also, $A=8$ because $c_5=A3,c_9=A9,c_3=13,c_4=23,c_{12}=59,c_{21}=73$. We have $D=7$ because $c_{11}=D9,c_7=19,c_9=29,c_{12}=59,c_9=89$. It follows from fact 2 that $c_{15}=c_{20}=71$. This is a contradiction.)
$c_2=C1$ and $c_{1}=31$.
(If $c_5=29$ and $c_2=13$, then $c_1,c_2,c_3,c_4,c_7,c_8,c_{21}$ are of the form $\circ 3$. From fact 1, this is impossible.
If $c_8=29$ and $c_2=13$, then $c_7,c_8,c_9,c_{11},c_{12}$ are of the form $\circ 9$. From fact 2, we have $c_{16}=53$, so $c_1,c_2,c_3,c_4,c_5,c_{16},c_{21}$ are of the form $\circ 3$. From fact 1, this is impossible.)
$B\not=1$.
(Suppose that $B=1$.
If $c_5=29$, then $c_{8}=A3,c_7=D3$. $$
\begin{array}{c|c|c|c|c}
31 & C1 & C3 & 23 & 29\\\hline
& \times & D3 & A3 & A9\\\hline
& \times & D9 & \times & 59 \\\hline
& & & \times & \\\hline
& & & & 13\end{array}
$$ We have $A=7,8$ because $c_8=A3,c_9=A9,c_{21}=13,c_4=23,c_{12}=59$. Also, $D=7,8$ because $c_7=D3,c_{11}=D9,c_{21}=13,c_{4}=23,c_{12}=59$. So, $C=4$ because $c_2=C1,c_3=C3,c_{21}=13,A3$ or $D3=73$.
If $c_{15}=19$, then $c_{19}=1G,c_{18}=4G$. $$
\begin{array}{c|c|c|c|c}
31 & C1 & C3 & 23 & 29\\\hline
& \times & D3 & A3 & A9\\\hline
& \times & D9 & \times & 59 \\\hline
& & 19 & \times & \\\hline
& 4G & 1G & & 13\end{array}
$$ So, $c_{14}=c_{19}=1G$. This is a contradiction. So, $c_{15}=71$ (this is because $p=19,71$ are the only possible primes satisfying $c_{15}=p$). Then, $c_1,c_2,c_{14},c_{15},c_{18},c_{19}$ are of the form $\circ 1$. From fact 3, this is impossible.
If $c_8=29$, then $c_5=A3,c_7=C9$. $$
\begin{array}{c|c|c|c|c}
31 & C1 & C3 & 23 & A3\\\hline
& \times & C9 & 29 & A9\\\hline
& \times & D9 & \times & 59 \\\hline
& & & \times & \\\hline
& & & & 13\end{array}
$$ So, $C=7$ because $c_2=C1,c_3=C3,c_7=C9,c_{21}=13$. Also, $A=8$ because $c_5=A3,c_9=A9,c_{21}=13,c_4=23,c_{12}=59,c_3=73$. We have $D=1$ because $c_{11}=D9,c_8=29,c_{12}=59,c_7=79,c_9=89$. It follows from fact 2 that $c_{11},c_{15},c_{19},c_{20},c_{21}$ are of the form $1\circ$. From fact 5, this is impossible.)
$c_8=29$.
(Suppose that $c_5=29$. We have $c_8=A3,c_7=D3$.
If $c_{20}=13$, then it follows from fact 1 that $c_{16}=B9$. $$
\begin{array}{c|c|c|c|c}
31 & C1 & C3 & 23 & 29\\\hline
& \times & D3 & A3 & A9\\\hline
& \times & D9 & \times & 59 \\\hline
& & & \times & B9 \\\hline
& & & 13 & B3\end{array}
$$ We have $A=7,8$ (because $c_8=A3,c_9=A9,c_{20}=13,c_4=23,c_{12}=59$), $B=7,8$ (because $c_{21}=B3,A_{16}=B9,c_{20}=13,c_4=23,c_{12}=59$), $D=7,8$ (because $c_7=D3,c_{11}=D9,c_{20}=13,c_4=23,c_{12}=59$). This is impossible. So, $c_{20}=B1$.
If $c_{16}=B9$, then we have $$
\begin{array}{c|c|c|c|c}
31 & C1 & C3 & 23 & 29\\\hline
& \times & D3 & A3 & A9\\\hline
& \times & D9 & \times & 59 \\\hline
& & & \times & B9 \\\hline
& & & B1 & B3\end{array}
$$ So, we have $A=1,7,8$ (because $c_8=A3,c_9=A9,c_4=23,c_{12}=59$), $B=1,7$ (because $c_{20}=B1,c_{21}=B3,c_{16}=B9,c_4=23,c_{12}=59$), $C=1,4,7$ (because $c_2=C1,c_3=C3$), $D=1,7,8$ (because $c_7=D3,c_{11}=D9,c_4=23,c_{12}=59$). Since $\{A,B,D\}=\{1,7,8\}$, we have $C=4$. Since $c_{19}=\circ 1$, we have $c_{18}=c_{2}=41$. This is a contradiction. So, $c_{16}=53$.
We have $c_{18}=41$. $$
\begin{array}{c|c|c|c|c}
31 & C1 & C3 & 23 & 29\\\hline
& \times & D3 & A3 & A9\\\hline
& \times & D9 & \times & 59 \\\hline
& & & \times & 53 \\\hline
& 41 & \circ 1 & B1 & B3\end{array}
$$So, $A=1,7,8$ (because $c_8=A3,c_9=A9,c_4=23,c_{12}=59$), $B=1,7$ (because $c_{20}=B1,c_{21}=B3,c_{18}=41$), $C=1,7$ (because $c_2=C1,c_3=C3,c_{18}=41$), $D=1,7,8$ (because $c_7=D3,c_{11}=D9,c_4=23,c_{12}=59$). We have $\{B,C\}=\{1,7\}$, so $A=8$. However, this is impossible since there is no such $D$.)
$c_5=A3,c_7=C9,c_{10}=97,c_6=37,c_{16}=53$.
(From fact 2, we have $c_{10}=97$ and $c_{16}=53$.)
$c_{20}=13$.
(Suppose that $c_{20}=B1$. Then, we have $c_{19}=E1,c_{18}=41$.
If $D\not=1$, then it follows from fact 2 that $c_{15}=D1$. Then, $c_1,c_2,c_{15},c_{18},c_{19},c_{20}$ are of the form $\circ 1$. From fact 3, this is impossible. So, $D=1$.
We have $$
\begin{array}{c|c|c|c|c}
31 & C1 & C3 & 23 & A3\\\hline
37 & \times & C9 & 29 & A9\\\hline
97 & \times & 19 & \times & 59 \\\hline
& & & \times & 53 \\\hline
& 41 & E1 & B1 & B3\end{array}
$$ So, $C=7,A=8,B=1,E=6$, but there is no prime $p$ such that $c_{15}=p$.)
$c_{19}=11$.
(We have $c_{19}=1E$ and $c_{18}=4E$.
Suppose that $E=7$. Then, $c_6,c_{10},c_{13},c_{14},c_{15},c_{17},c_{18},c_{19}$ are of the form $\circ 7$. From fact 4, this is impossible.)
$c_{18}=41,C=7,A=8,B=4,D=1,c_{14}=47$.
(Suppose that $c_{17}=47$. Then, $c_1,c_2,c_{14},c_{15},c_{18},c_{19}$ are of the form $\circ 1$. From fact 3, this is impossible.)
$c_{15}=17,c_{13}=67,c_{17}=61$.