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I came across a question that I couldn't figure out. It was:

What is the value of all the letters in this following cryptarithm? $$\begin{array}{ccccccc} &&&&T&H&E\\ +&&&B&E&S&T\\ &S&Y&S&T&E&M\\ \hline &M&E&T&R&I&C\\ \end{array}$$

The problems

I can't figure out any of the letters. It is just like a bunch of letters together to form an equation.

What I know

The TET column and ETM column has different answers, an R and a C. In the 2 columns are 2 common letters: an E and a T. So, we can justify that the difference between E and T(or T and E) equals the difference between R and C(or C and R).

Another thing I know is that the same TET column and the HSE column next to it are different and has one common letter, E. So the 2 columns will become a 2T column and an H+S column. Therefore, 2T's does not equal H+S as the answers are different.

I have been staring at this question, my head is blank for an hour or so. Can I please have some help?

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2 Answers 2

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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.

  1. If $T=1$, then $R=2$ and $B+M=12$.

    • If $(B,M)=(8,4)$, then $S=M-1=3$ and $(E\,T\,M\,|\,C)$ gives $$C=(E+T+M)\text{ mod }10=5\,.$$ Thus, $$(Y,B,C,M,S,R,T,E)=(9,8,5,4,3,2,1,0)\,.$$ Thus, $(H\,S\,E\,|\,I)$ yields $$I=(H+S)\text{ mod }10=(H+3)\text{ mod }10\,.$$ This cannot be fulfilled with $\{I,H\}=\{6,7\}$.
    • If $(B,M)=(7,5)$, then $S=M-1=4$ and $(E\,T\,M\,|\,C)$ gives $$C=(E+T+M)\text{ mod }10=6\,.$$ Thus, $$(Y,B,C,M,S,R,T,E)=(9,7,6,5,4,2,1,0)\,.$$ Thus, $(H\,S\,E\,|\,I)$ yields $$I=(H+S)\text{ mod }10=(H+4)\text{ mod }10\,.$$ This cannot be fulfilled with $\{I,H\}=\{3,8\}$.
    • If $(B,M)=(5,7)$, then $S=M-1=6$ and $(E\,T\,M\,|\,C)$ gives $$C=(E+T+M)\text{ mod }10=8\,.$$ Thus, $$(Y,C,M,S,B,R,T,E)=(9,8,7,6,5,2,1,0)\,.$$ Thus, $(H\,S\,E\,|\,I)$ yields $$I=(H+S)\text{ mod }10=(H+6)\text{ mod }10\,.$$ This cannot be fulfilled with $\{I,H\}=\{3,4\}$.
    • If $(B,M)=(4,8)$, then $S=M-1=7$ and $(E\,T\,M\,|\,C)$ gives $$C=(E+T+M)\text{ mod }10=9\,,$$ which is contradiction ($Y=9$ already).
  2. If $T=2$, then $R=4$ and $B+M=13$.

    • If $(B,M)=(8,5)$, then $S=M-1=4=R$, which is a contradiction.
    • If $(B,M)=(7,6)$, then $S=M-1=5$ and $(E\,T\,M\,|\,C)$ gives $$C=E+T+M=8\,.$$ Thus, $$(Y,C,B,M,S,R,T,E)=(9,8,7,6,5,4,2,0)\,,$$ whence $(H\,S\,E\,|\,I)$ yields $$I=(H+S)\text{ mod }10=(H+5)\text{ mod }10\,.$$ This cannot be fulfilled with $\{I,H\}=\{1,3\}$.
    • If $(B,M)=(6,7)$, then $S=M-1=6=B$, which is a contradiction.
    • If $(B,M)=(5,8)$, then $S=M-1=7$ and $(E\,T\,M\,|\,C)$ gives $$C=(E+T+M)\text{ mod }10=0\,,$$ which is contradiction ($E=0$ already).
  3. 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.

  4. 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 $T=3$.

  1. 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).

  2. 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.

  3. 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.

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  • $\begingroup$ This is very helpful @Batominovski. Thank you for helping me! $\endgroup$
    – Stevo
    Apr 12, 2020 at 10:41
  • $\begingroup$ Can you simplify your answer for Year 5's? $\endgroup$
    – Stevo
    Apr 12, 2020 at 11:47
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    $\begingroup$ @102152111 Unfortunately, I have lost the energy to deal with this problem. It is too long. Expanding this will make the solution probably at least twice longer. Let's hope someone will come up with a shorter solution then. $\endgroup$ Apr 12, 2020 at 11:49
  • $\begingroup$ No worries. Thanks anyway $\endgroup$
    – Stevo
    Apr 12, 2020 at 11:51
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This is a cryptarithmic puzzle. A brute-force search shows that this is the unique solution: $$\begin{array}{ccccccc} &&&&1&8&0\\ &&&7&0&4&1\\ +&4&9&4&1&0&5\\ \hline &5&0&1&3&2&6 \end{array}$$ So $THEBSYMRIC=1807495326$.

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