4
$\begingroup$
  • Every question mark and every letter stands for a base-10 digit
  • Different letters stand for different digits
  • Question marks are placeholders and stand for arbitrary digits (such digits also may occur for letters)
  • Leading digits are always non-zero.

            CATS
           * EAT
          -------
           ?????
          ?????
         ?????
         --------
         ?MOUSES
    

Which digit does each letter represent?
(Please present the full analysis how these digits can be determined.)

$\endgroup$
6
$\begingroup$

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 
$\endgroup$
  • $\begingroup$ Pah! Just about to post my answer when I see you've beaten me too it. Yours is better formatted anyway. Well done! $\endgroup$ – Gordon K Mar 22 '16 at 16:14
  • $\begingroup$ @GordonK I always struggle with the format. It always seems like such a run on logic sentence that I don't know if it is clear enough or not. $\endgroup$ – Trenin Mar 22 '16 at 17:23
  • $\begingroup$ @GordonK Yep, I kind of wish background colour would change rather than just that little grey box - I didn't notice there was an answer and just did it too; second time I've done that. I got the same results but kept possibilities in a mutating set rather than writing out the hypotheses as Trenin did. $\endgroup$ – Jonathan Allan Mar 23 '16 at 2:07

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