How many ways can the digits $0,1,2, \ldots, 9$ be written in place of the letters in the following schema to get a correct addition? (0 cannot be the first digit in any row; the answer can be given without writing out all the possibilities.)

$$
\begin{array}{r}
A \\
B C \\
D E F \\
\hline C H J K
\end{array}
$$
There are ten letters, and a different digit must be written in place of each. The sum is only four-digit if $D=9$, because the sum of the digits in the last column is less than 30, so it can carry at most 2 to the tens column; here, however, the maximum is $9+8$, so at most 1 carry is produced. From this, we also see that the first two digits of the sum are $G=1, H=0$.

Since 0, 1, and 9 cannot reappear, the sum of the last column is at least $2+3+4=9$, and at most $8+7+6=21$. Furthermore, $K$ must also be a new digit, so the sum cannot be $9,10,11$, nor 21, 20, 19, thus

$$
12 \leqq A+C+F \leqq 18
$$

so a remainder of 1 is always carried to the tens column:

$$
A+C+F=10+K, \quad \text { and } \quad 1+B+E=10+J
$$

The sum of the remaining digits, regardless of their order,

$$
A+B+C+E+F+J+K=2+3+4+5+6+7+8=35
$$

subtracting the sum of the above two equations

$$
J+K=8, \quad \text { so } \quad J=6,5,3, \quad \text { or } \quad 2 .
$$

If $J=6$, then from (1) $B+E=15$, and only $B=7, E=8$ satisfies this - since the last two columns can be filled with decreasing digits for now, as no digit can be 0, then $K=2$, and the remaining digits $A$, $C, F$ must necessarily satisfy (1) according to (2) and (3).

If $J=5$, similarly $B=6, E=8$; if $J=3$, then $B=4, E=8$.

Finally, if $J=2$, then $K=6$, the tens column can be filled in two ways, with $B=4$ or 3. Thus, essentially there are five different solutions:

| 3 | 2 | 2 | 3 | 4 |
| ---: | ---: | ---: | ---: | ---: |
| 74 | 64 | 46 | 45 | 35 |
| 985 | $\frac{987}{1053}$ | $\frac{987}{1035}$ | $\frac{978}{1026}$ | $\frac{987}{1026}$ |

II. The two digits in the tens column can always be written in 2 different orders, and the three digits in the units column can be written in $3 \cdot 2=6$ orders, because the position of $A$ can be filled in 3 ways, and the remaining two positions can always be filled with the 2 remaining digits in 2 ways. Each order of the tens column combined with each order of the units column results in a new arrangement, so from each of the above solutions, $2 \cdot 6=12$ order variations can be formed, thus the number of possible entries is $5 \cdot 12=60$.

Note. We can also arrive at equation (3) based on the so-called 9's test. Let the sum of the digits above the line be $f$, and below the line be $a$, so $a$ differs from $f$ by some multiple of 9: $f=a+9k$, where $k$ is an integer. On the other hand, $f+a=0+1+\ldots+8+9=45$, so $2a+9k=9 \cdot 5$, thus $a$ is divisible by 9, but after using 0, 1, and 9, it cannot even reach 18, so $a=9$.