13. There are two equations:

Good + Good $=$ Wonderful, Wonderful $\times$ GoodGood $\times$ ReallyGood $=$ WonderfulProblemProblemWonderful, where each Chinese character represents a digit from $0 \sim 9$, the same character represents the same digit, and different characters represent different digits. Therefore, the number of all factors of the four-digit number "WonderfulProblemProblemWonderful" is $\qquad$
13.16.

From the addition formula, we get “好” $<5$, “妙” $\neq 0$, so “好” $=1$, “妙” $=2$ or “好” $=2$, “妙” $=4$ or “好” $=3$, “妙” $=6$ or “好” $=4$, “妙” $=8$. Clearly, the middle two scenarios do not satisfy the multiplication formula, so it can only be:
(1) “好” $=1$, “妙” $=2$, thus the multiplication formula becomes
$2 \times 11 \times($ 真 $\times 10+1)=2002+$ 题 $\times 110$, i.e., 真 $\times 10+1=91+$ 题 $\times 5$.
The left side $\leqslant 91$, the right side $\geqslant 91$, so both sides equal 91. Therefore, “真” $=9$, “题” $=0$, “妙题题妙” $=2002$.
(2) “好” $=4$, “妙” $=8$, the multiplication formula is
$8 \times 44 \times($ 真 $\times 10+4)=8008+$ 题 $\times 110$,

i.e., $704+1760 \times$ 真 $=4004+$ 题 $\times 55$.
In the range $0 \sim 9$, only “真” $=2$, “题” $=4$ satisfy the above equation, but in this case, “好” and “题” represent the same digit, which contradicts the problem statement.
Therefore, the four-digit number “妙题题妙” has a unique solution 2002.
Since $2002=2 \times 7 \times 11 \times 13$, the number of all divisors of 2002 is $2^{4}=16$.