cryptarithmpuzzle
Given multiplication is
$$\begin{array}{cccccc}
& & & P & E & N \\
& & & I & N & K \\\hline
& & L & K & P & R \\
& I & R & T & N & \\
E & A & K & N & & \\\hline
L & E & T & T & E & R
\end{array}$$
how to find the values of the letters?
$$2S+Y=1+10\alpha$$ $$A+R+\alpha=10$$ $$M+2R+1=D+10\gamma$$ $$2E+X+\gamma=A+10\epsilon$$ $$H+M+\epsilon=10R+E$$ $\alpha,\gamma,\epsilon \le 2$ and, since $R\ne 0$, $R$ must be $1$. From $2S+Y=1+10\alpha$ we have $\alpha=2$ and therefore $A=7$ or $\alpha=1$ and therefore $A=8$. We will deal with these possibilities separately.
If $(A,R)=(7,1)$ and $(S,Y)=(9,3),(8,5)\text { or }(6,9).$
$$(1) M+3=D+10\gamma$$ $$(2) 2E+X+\gamma=7+10\epsilon$$ $$(3) H+M+\epsilon=10+E$$
The case $\epsilon=\gamma=0$
From (2) we have $(E,X)=(2,3).$ From (3)-(1) we have $\{D,H\}=\{6,9\}.$ Then $M=D-3= \text{3 or 6}$, a contradiction.
The case $\epsilon=0,\gamma=1$
From $M=D+7$, we have $(D,M)=(2,9).$ Then $X$ cannot be $0$ and so $2E+X=6$ gives $(E,X)=(0,6).$ Then $H=1+E=1$, a contradiction.
If $\epsilon\ne 0$
Adding equations (2) and (3), $E+H+M+X+\gamma=17+9\epsilon\ge 26$. However, $E+H+M+X\le 9+6+5+4=24$ and $\gamma\le1$ so this is impossible.
If $(A,R)=(8,1)$ and $(S,Y)=(4,3),(3,5)\text { or }(2,7).$
$$(1) M+3=D+10\gamma$$ $$(2) 2E+X+\gamma=8+10\epsilon$$ $$(3) H+M+\epsilon=10+E$$
The case $\epsilon=\gamma=0$
From (2) we have $(E,X)=(4,0)\text { or }(2,4).$ From (3)-(1) we have $E=2,\{D,H\}=\{6,9\}.$ Then$(S,Y)=(3,5)$ and $M=D-3= \text{3 or 6}$, a contradiction.
The case $\epsilon=0,\gamma=1$
From $M=D+7$, we have $(D,M)=(0,7) \text { or } (2,9).$ Then $(S,Y)=(4,3)\text{ or } (3,5).$ Then $0$ and $3$ are already assigned and so $2E+X=7$ gives $(E,X)=(0,7).$ Then $H+M=10$ and $H=1$ or $3$, a contradiction.
If $\epsilon\ne 0$
Adding equations (2) and (3), $E+H+M+X+\gamma=18+9\epsilon$. Therefore $\epsilon=1$.
Since $8$ is already assigned either $$\{E,H,M,X\}=\{9,7,6,5\},\gamma=0 \text { or } \{E,H,M,X\}=\{9,7,6,4\},\gamma=1. $$ In the first case, $2E+X=18$ and $X$ is even, then $X=6$ and $E=X$, a contradiction.
In the second case, $2E+X=17$ and $X$ is odd, then
$X=9,E=4,M=D+7$. Therefore $D=0,H=6,M=7.$ Finally, the only possibility for $S$ and $Y$ is $S=3,Y=5$.
Assume that all variables are distinct digits. Then, by inspecting the columns $(\_\,\_\,S\,|\,M)$ and $(\_\,\_\,Y\,|\,E)$, we clearly have $M=S+1$ and $(Y+1)\operatorname{mod}10=E$. Since there must be a carry-over from $Y+1$, we must have $E=(Y+1)-10$. Thus, $E=Y-9$. This shows that $Y=9$ and $E=0$. Now, the column $(\_\,B\,S\,|\,T)$ gives either $$(B+S)-10=T\text{ or }(B+S+1)-10=T$$ (recalling the carry-over to the column $(\_\,\_\,Y\,|\,E)$). Since $9$ is taken by $Y$ and $M=S+1$, we get $$T\leq (B+S+1)-10=(B+M)-10\leq (8+7)-10=5\,.$$
If $T=5$, then we must have $\{B,M\}=\{8,7\}$. As $S=M-1$, we get $$(Y,B,M,S,T,E)=(9,8,7,6,5,0)\,.$$ By considering the column $(E\,T\,M\,|\,C)$, we conclude $$C=(E+T+M)\operatorname{mod}10=(0+5+7)\operatorname{mod}10=2\,.$$ From the column $(H\,S\,E\,|\,I)$, we obtain (recalling the carry-over from the column $(E\,T\,M\,|\,C)$) $$I=(H+S+E+1)\operatorname{mod}10=(H+6+0+1)\operatorname{mod}10=(H+7)\operatorname{mod}10\,.$$ The only possible values of $H$ are $1$, $2$, $3$, and $4$; however, none of these values will make $I$ a distinct digit from previously known digits. Thus, $T=5$ is false.
We have proven that $T<5$. Because $E=0$, the carry-over to $(T\,E\,T\,|\,R)$ from $(H\,S\,E\,|\,I)$ is at most $1$. This means either $$R=2T\text{ or }R=2T+1\,.$$ Recall from $(\_\,B\,S\,|\,T)$ that $B+S-10=T$, or $$B+M=B+(S+1)=T+11\,.$$
We first assume that $R=2T$. We have the following cases.
If $T=1$, then $R=2$ and $B+M=12$.
If $T=2$, then $R=4$ and $B+M=13$.
If $T=3$, then $R=6$ and $B+M=14$. Since $B$ and $M$ are at most $8$ and unequal, we must have $$B=6=R\text{ or }M=6=R\,,$$ which is a contradiction.
If $T=4$, then $R=8$ and $B+M=15$. As $B$ and $M$ are now at most $7$, $$B+M\leq 14<15\,,$$ which is a contradiction.
Ergo, $R=2T+1$ must be the case. Since $R<9$ and $T>0$, we see that $T=1$, $T=2$, or $T=3$.
If $T=3$, then $R=7$ and $B+M=14$. Since $S=M-1$ cannot equal $R=7$, we end up with $$(Y,B,R,M,S,T,E)=(9,8,7,6,5,3,0)\,.$$ Consequently, $(E\,T\,M\,|\,C)$ gives $$C=(E+T+M)\text{ mod }10=9\,,$$ which is a contradiction ($Y=9$ already).
If $T=2$, then $R=5$ and $B+M=13$. Clearly, $M=13-B\geq 13-8=5$. As $M\neq R=5$ and $S=M-1\neq R=5$, we must have $M\geq 7$.
If $M=7$, then $B=13-M$ and $S=M-1=6$, which is a contradiction.
If $M=8$, then $B=13-M=5=R$, which is again a contradiction.
If $T=1$, then $R=3$ and $B+M=12$. Consequently, $(E\,T\,M\,|\,C)$ gives $$C=(E+T+M)\text{ mod }10=M+1\,.$$ As $C\leq 8$, we get $M\leq 7$.
If $(B,M)=(8,4)$, then $S=M-1=3=R$, which is a contradiction.
If $(B,M)=(7,5)$, then $S=M-1=4$ and $C=M+1=6$. This gives $$(Y,B,C,M,S,R,T,E)=(9,7,6,5,4,3,1,0)\,.$$ Thus, $(H\,S\,E\,|\,I)$ yields $$I=(H+S)\text{ mod }10=(H+4)\text{ mod }10\,.$$ This can be fulfilled only by $(I,H)=(2,8)$. Thus, we have a unique solution $$(Y,H,B,C,M,S,R,I,T,E)=(9,8,7,6,5,4,3,2,1,0)\,.$$
Epilogue. Without the requirement that the digits must be distinct, there are many other solutions. Via computer search, there are $7145$ solutions with $T$, $B$, $S$, and $M$ being positive (so that $THE$, $BEST$, $SYSTEM$, and $METRIC$ are $3$-, $4$-, $6$-, and $6$-digit positive integers). Without the positivity requirements (i.e., $T$, $B$, $S$, and $M$ may be $0$), there are $9900$ solutions.
Best Answer
Start from N*N ends in N, so N must be 5 or 6 (can't be 0 or 1 from the second line in the sum). Then I*N ends in N (and I !=1) says N is 5 and I is odd. Then K is even and R is 0. Continue