# A condemning alphametic

• 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.

         SIN
* IN
-------
????
U?S
-------
?EVIL


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

## 2 Answers

The answer is

There is one valid solution which is $S=4, I=2, N=7, U=8, E=1, V=5, L=9$

Formatting as above

427
*   27
---------
2989
854
---------
11529

Proof

First we note that $N \times N \equiv L (\bmod 10)$ so immediately we cannot have $N=0,1,5,6$ as then $L=N$ which is not permitted. Also important is $I \times N \equiv S (\bmod 10)$ (1)
We can thus consider the problem for different values of $N$. In the following, let $M_i$ represent the $i$th digit in the first multiplication (reading left to right)

$N=2 \Rightarrow L=4 \Rightarrow M_3 + S \equiv I (\bmod 10)$ and we must have $M_3 = S$ but then from (1) above that means that $4S \equiv 2I \equiv S (\bmod 10)$ which means $S=0$ and this cannot be since it is a leading digit.

$N=3 \Rightarrow L=9 \Rightarrow M_3 = S$ and using (1) we must have $6S \equiv 3I \equiv S (\bmod 10)$ which allows $S=2,4,6,8$ paired with $I=4,8,2,6$ respectively. Clearly, since the second multiplication has one less digit than the first, we must have $I < N$ so we can rule out all but the third of these possibilities but then $623 \times 2$ has four digits instead of three so this won't work.

$N=4 \Rightarrow L=6 \Rightarrow M_3 = S+1$ which means that $8S+4 \equiv 4I \equiv S (\bmod 10)$ which means $S=8$ and $I=7$ but this doesn't work since we must have $I<N$.

$N=7 \Rightarrow L=9 \Rightarrow M_3 \equiv S+4 (\bmod 10)$ which means that $14S+28 \equiv 7I \equiv S(\bmod 10)$ or more simply $3S \equiv 2(\bmod 10)$ which gives $S=4$ and $I=2$. This gives a valid solution with additionally $U=8, E=1, V=5$

$N=8 \Rightarrow L=4 \Rightarrow M_3 \equiv S+6 (\bmod 10)$ which means that $16S + 48 \equiv 8I \equiv S (\bmod 10)$ or more simply $5S \equiv 2 (\bmod 10)$ which has no solutions.

Finally, $N=9 \Rightarrow L=1 \rightarrow M_3 \equiv S+8 (\bmod 10)$ which gives us $18S+72 \equiv 9I \equiv S(\bmod 10)$ or more neatly $7S \equiv 8 (\bmod 10)$ which gives $S=4$ and $I=6$ but this gives four digits in the second multiplication instead of three.

Having checked all the cases, we find there is only one valid solution.

Solution:

$S=4, I=2, N=7, U=8, E=1, V=5, L=9$

Proof:

I is not 1, since $SIN*I = U?S$
IS + carriage from II < 10. So I * S < 10.
This means
[S, I] can be one of [2,3], [2,4], [3,2], [4,2].
Since N*N ends in L it means N cannot be one of 0,1,5,6.

For SIN = 23N, the biggest possible value is
$SIN = 239$
239 *
39
---
9321 - wrong. there are only 4 digits
This is not good
For $SIN = 32N$ the biggest possible value is 329 * 29 ----- 9541 - wrong. there are only 4 digits Again, not good. so [S, I] can be [2, 4] or [4,2]
For $SIN = 243$

243*
43
---
729 - this should be 4 digits
???
-----
???? -
For $SIN = 247$

247*
47
---
1729
988
-----
11609- this results in I = 0 so it's wrong

For $SIN = 248$

248*
48
---
1984
992
-----
11904 - this results in L = I so it's wrong

FOR $SIN=249$

249*
49
---
2241
996
-----
12201 - this results in E = V so it's wrong

For $SIN = 423$

423*
23
---
1269
846
-----
9729 - it should have 5 digits so it's wrong

For $SIN = 427$

427*
27
---
2989
854
-----
11529 - we have a winner.

For $SIN = 428$

428*
28
---
3424
856
-----
11984 - this results in N = I which is wrong.

For $SIN = 429$

429*
29
---
3861
858 - this means U = S so it's wrong
-----
?????