# An alphametic puzzle about Charles Darwin

Charles Darwin was born on February 12, 1809.

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

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

I used this application to solve this puzzle:

Solution 2
00:00:04.2871214

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

• The first solution does not work because it assigns the digit 6 to both N and C, which is not permitted. The second solution is valid. That said, posting a solution found by a computer program is generally frowned upon for this type of puzzle. This is particularly true if you are not the author of said program. Commented Feb 12, 2023 at 20:16
• ok only one solution. I am the developer but. my app has options and the user can miss to use all possibilities in this case the user is the developer. Thank you for your notice to me Commented Feb 12, 2023 at 20:46
• I have edited your answer to remove the invalid solution and to contain the other in a spoiler block. Note the use of 'pre' tags for preformatted text within the spoiler, and that the spoiler tag is needed on every line. I do not want to discourage the development of your app or your participation in this forum, but please understand what is expected in an answer. For this and most other logic-based puzzles, an explanation of the logical deductions needed to arrive at the solution is preferred. Commented Feb 12, 2023 at 21:44
• thank you for your kindness. in this case what I can do is to add the filters I used in the application to achieve the solution. because the application is to resolve standard Alphametic puzzles with filters for none standard puzzles. if that ok to this forum I'll in if not I'll out friendly. Thank you very much Commented Feb 13, 2023 at 7:35