First, we see that every product is a 5 digit number, so we know that $E,A,T \notin \{0,1\}$.
Now, we also see in the ones column that $S \times T \implies S$. This works in the following cases:
- $S=0$
- $S\in \{2,4,8\}, T=6$
- $S=5, T \in \{3,7,9\}$
Assume $S=0$
In the tens column, we have $T^2 \implies E$. Thus,
$$T\in\{2,3,4,5,6,7,8,9\}$$
$$E\in\{4,9,6,5,6,9,4,1\}$$
We will denote $c_2$ to be the carry over of $T^2$ into the hundreds column.
From the hundreds column, we see that
$$c_2 + T \times A + T \times A = 2 \times (T \times A) + c_2 \implies 0$$
If $T=4, E=6$, then $c_2=1$ and there is no solution since we would need $2 \times (T \times A)$ to be 9, which is odd, so this possibility can be ruled out.
Of the other possibilities, some are invalid and can be removed. We are left with:
$$T\in\{2,3,7,8\}$$
$$E\in\{4,9,9,4\}$$
$$c_2\in\{0,0,4,6\}$$
These take a bit of work.
Assume $T=2,E=4, c_2=0, A=5$
All that is missing is $C$, so lets try the options:
- $1520 \times 452 = 687040$, but result needed to be one more digit
- $3520 \times 452 = 1591040 \implies M=A=5$
- $6520 \times 452 = 2947040 \implies O=E=4$
- $7520 \times 452 = 3399040 \implies O=U=9$
- $8520 \times 452 = 3851040 \implies O=A=5$
- $9520 \times 452 = 4303040 \implies M=U=3$
Assume $T=3, E=9, c_2=0, A=5$
Running through the options for $C$:
- $1530 \times 953 = 1458090 \implies O=A=5$
- $2530 \times 953 = 2411090 \implies O=U=1$
- $4530 \times 953 = 4317090 \implies M=T=3$
- $6530 \times 953 = 6223090 \implies M=O=2$
- $7530 \times 953 = 7176090 \implies C=O=7$
- $8530 \times 953 = 8129090 \implies U=E=9$
Assume $T=7, E=9, c_2=4, A=4$
Running through the options for $C$:
- $1470 \times 947 = 1392090 \implies O=E=9$
- $2470 \times 947 = 2339090 \implies M=O=3$
- $3470 \times 947 = 3286090$ Solution 1!
- $5470 \times 947 = 5180090 \implies U=S=0$
- $6470 \times 947 = 6127090 \implies U=T=7$
- $8470 \times 947 = 8021090 \implies M=S=0$
Assume $T=8, E=4, c_2=6, A=9$
Running through the options for $C$:
- $1980 \times 498 = 986040$, but result needed to be one more digit
- $2980 \times 498 = 1484040 \implies M=E=4$
- $3980 \times 498 = 1982040 \implies M=A=9$
- $5980 \times 498 = 2978040 \implies M=A=9$
- $6980 \times 498 = 3476040 \implies M=E=4$
- $7980 \times 498 = 3974040 \implies M=A=9$
Assume $S=2, T=6, c_1=1$
In the tens column, we have
$$c_1 + T^2 + A \times S = 1 + 36 + 2A = 37 + 2A \implies E$$ Thus, $E$ is odd.
If $A=3$ then $E=3$. If $A=7$ then $E=1$. So both those can be removed.
In the hundreds column, we have
$$c_2+ A \times T + A \times T + E \times S = c_2 + 12A + 2E \implies S$$
If $A=5$, then from the tens, we know $E=7$. But from the hundreds, $c_2=4$ and $4+60+14 =78 \implies S=8$.
If $A=8$, then from the tens, $E=3$. From the hundreds, $c_4=5$ and $5+96+6=107 \implies S=7$.
If $A=9$, then from the tens, $E=5$. From the hundreds, $c_4=5$ and $5+108+10=123 \implies S=3$.
All that remains to check is when $A=4, E=5$.
Assume $A=4, E=5$
Running through the possibilities for $C$ gives:
- $1462 \times 546 = 798252$, but result needed to be one more digit
- $3462 \times 546 = 1890252$, Solution 2!
- $7462 \times 546 = 4074252 \implies O=C=7$
- $8462 \times 546 = 4620252 \implies O=S=2$
- $9462 \times 546 = 5166252 \implies O=U=1$
Assume $S=4, T=6, c_1=2$
In the tens column, we have
$$c_1 + T^2 + A \times S = 2 + 36 + 4A = 38 + 4A \implies E$$ Thus, $E$ is even.
If $A=2$ , then $E=T=6$.
If $A=8$, then $E=0$.
Assume $S=8, T=6, c_1=4$
In the tens column, we have
$$c_1 + T^2 + A \times S = 4 + 36 + 8A = 40 + 8A \implies E$$ Thus, $E$ is even.
If $A=2$, then $E=T=6$.
Thus $A=4, E=2, c_2=7$.
In the hundreds column, we have
$$c_2+ A \times T + A \times T + E \times S = 7 + 24+24 + 16 = 61 \implies S=1$$
Assume $S=5, T \in \{3,7,9\}$
From the tens, we know that
$$c_1 + T^2 + A \times S = c_1 + T^2 + 5A \implies E$$
If $T=3$, then $c_1=1$, so $1+9+5A =10 + 5A \implies E$. That makes $E\in\{0,5\}$ which is not allowed.
If $T=9$, then $c_1=4$, so $4+81+5A =85 + 5A \implies E$. That makes $E\in\{0,5\}$ which is not allowed.
Thus $T=7$ and $c_1=3$, so $3+49+5A=52+5A \implies E$. Thus $E\in\{2,7\}$ but 7 is already taken so $E=2$ and $A$ is even.
From the hundreds, we know that
$$c_2 + T \times A + T \times A + E \times S = c_2 + 14A + 10 \implies 5$$
If $A=2$, then $c_2=6$, so $6+28+10=46 \implies S=6$.
If $A=4$, then $c_2=7$, so $7+56+10=73 \implies S=3$.
If $A=6$, then $c_2=8$, so $8+84+10=102 \implies S=2$.
If $S=8$, then $c_2=9$, so $9+112+10=131 \implies S=1$.
Solutions
Thus, there are 2 solutions.
CATSEMOU=34709286
3470
* 947
----
24290
13880
31230
-------
3286090
And
CATSEMOU=34625890
3462
* 546
----
20772
13848
17310
-------
1890252