An alphametic puzzle about Charles Darwin

Question

Charles Darwin was born on February 12, 1809.

https://en.wikipedia.org/wiki/Charles_Darwin

Answer 1

Non-computer solution

$$(W,I,E,N,S,D,P,C,A,R) = (0,1,5,3,8,7,9,2,4,6)$$

Solution Path

Immediately from the way the division is performed we can see that $SPEC$ is divisible by $12$ and $12 \times WI$ is less than $IE$ which makes $W=0$.
Given that $12 \times I$ is less than $IE$, it must be that $I = 1,2,3$ or $4$.

If $I=4$ then $E$ must be $9$ so that $IE-(I\times 4)$ leaves a remainder. However, this would make $N=1$ and $S=4=I$, which is not allowed.

If $I=3$, then $E$ must be $7,8$ or $9$.
$E=7$ will leave a remainder of $1$ making $N=1$ and $S=4$. We'll call this case (i).
$E=8$ will leave a remainder of $2$ making $N=2$ and $S=6$. We'll call this case (ii).
$E=9$ will leave a remainder of $3$ making $N=3=I$, which is not allowed.

If $I=2$ then $E$ must be $5,6,7,8$ or $9$.
$E=5$ will leave a remainder of $1$ making $N=1$ and $S=4$. We'll call this case (iii).
$E=6$ leaves a remainder of $2$ making $N=2=I$, which is not allowed.
$E=7$ leaves a remainder of $3$ making $N=3$ and $S=8$. We'll call this case (iv).
$E=8$ leaves a remainder of $4$ but there is no value of $S$ which will yield $2$ more than a multiple of $12$ so we can disregard this.
$E=9$ leaves a remainder of $5$ but this would make $S=0=W$, which is not allowed.

$I=1$ puts $E=3,4,5,6,7,8,9$ but the previous analysis makes these cases easier to break down. In particular,
$E=3 \Rightarrow N=1=I$, which is not allowed.
$E=4 \Rightarrow N=2, S=6$, case (v)
$E=5 \Rightarrow N=3, S=8$, case (vi)
$E=6$ gives remainder $4$ which won't work (as before).
$E=7 \Rightarrow S=0=W$, which is not allowed.
$E=8 \Rightarrow N=5, S=2$, case (vii)
$E=9 \Rightarrow N=6, S=4$, case (viii)

Now there are eight cases but $S=2,4,6$ or $8$.
If $S=2$ then $D$ must be $1$ but this rules out case (vii) where $I=1$.
If $S=4$ then $D$ must be $3$ which rules out case (i) because $I=3$. This leaves six cases.

Case (ii) has $(I,E,N,S) = (3,8,2,6)$ and $D=5$, necessarily. Given that $D\times 12$ has second digit $0$, the remainder from the first calculation will be $P$. This means that that $PEC$ is divisible by $12$ with $E=8$ and $P,C$ chosen from $1,4,7,9$. However, $C$ must be even and since $84$ is already divisible by $12$ but $100$ isn't, there is no $P$ which will make $PEC$ divisible by $12$. Hence no solutions in this branch.

Case (iii) has $(I,E,N,S) = (2,5,1,4)$ and $D=3$, necessarily. To get a 1-digit remainder in the first calculation we must have $P < 6$ but this doesn't work since all digits are taken so no solution in this branch either.

Case (iv) has $(I,E,N,S) = (2,7,3,8)$. Here we can have $D=6$ or $D=7$.
$D=6$ must mean $P=1$ which leaves remainder $9$ after the first calculation. Then $PEC = 97C$ is divisible by $12$ which must make $C=2=I$, which is not allowed.
$D=7=E$ is also not allowed so no solution in this branch.

Case (v) has $(I,E,N,S) = (1,4,2,6)$ and $D=5$ necessarily. Here, the remainder from the first calculation will be $P$ and so $PEC$ is divisible by $12$ with $E=4$ and $P,C$ chosen from $3,7,8,9$. $C$ must be even so is $8$ and since $48$ is divisible by $12$ and $100$ isn't, there is no solution that works in this branch.

Case (vi) has $(I,E,N,S) = (1,5,3,8)$ and $D=6$ or $D=7$.
$D=6$ makes $P=1=I$ so this won't work.
$D=7$ means $P$ is greater than $4$ so is either $6$ or $9$.
$P=6$ gives remainder $2$ after the first calculation which means $25C$ is divisible by $12$ and so $C=2$. This, however makes $A=2=C$ so doesn't work.
$P=9$ gives remainder $5$ after the first calculation which means $55C$ is divisible by $12$ which means $C=2$. Then $A=4$ and $R=6$ which works! so overall we have $$(W,I,E,N,S,D,P,C,A,R) = (0,1,5,3,8,7,9,2,4,6)$$

Case (viii) has $(I,E,N,S) = (2,5,1,4)$ which makes $D=3$, necessarily.
Then $P<6$ to leave a 1-digit remainder in the first calculation but this is not possible since all digits are already taken in this range.

Overall just one solution $$(W,I,E,N,S,D,P,C,A,R) = (0,1,5,3,8,7,9,2,4,6)$$

The division computation appears as follows

Answer 2

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Solution 2
00:00:04.2871214

    746013
   -------
12|8952158
   84
   --
    55
    48
    --
     72
     72
     --
       15
       12
       --
        38
        36
        --
         2