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This puzzle was found from the Hot Network Questions on the right. A repost of a question in a way (from Mathematica SE), but I was wondering if the following puzzle could be done:

the puzzle

A quick translation:

Can anyone solve this? It's so hard!

Fill in the numbers $1–9$ in the boxes, the numbers cannot repeat.

It has already been solved in the given link, the answers being (hidden for those who wish to try and solve it):

$17\times4=68+25=93$

The answer to the Mathematica question stated there's a way to solve it without brute-forcing, but I didn't understand it too well. How would one—if possible—solve this without using too complex softwares?

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  • $\begingroup$ One way to get some nontrivial information is to consider everything modulo 9 and modulo 11. $\endgroup$ – Greg Martin Feb 22 '16 at 6:39
  • $\begingroup$ If you want to solve without using "too complex software" you can use a very simple on-line tool that I wrote years ago to solve and create cryptarithms. In this webpage, on the right panel you can enter $AB*C=DE$ and $DE+FG=HI$ one in each line, then hit "Solve". You get $4$ solutions, among which only one does not use the digit "$0$". $\endgroup$ – Giovanni Resta Feb 22 '16 at 9:40
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You can reduce the effort to a level that can be handled without a computer.

Let's write the calculation as

\begin{array}{cc} &A&B\\ \times&&C\\\hline &D&E\\ +&F&G\\\hline &H&I \end{array}

There are strong restrictions on $A$ and $C$, since their product must be single-digit. We can't have $C=1$, since otherwise $DE=AB$. So either $A=1$, or $(A,C)$ must be one of $(2,3)$, $(2,4)$, $(3,2)$ or $(4,2)$. We can exclude $(2,4)$ and $(4,2)$: This implies $D=8$, $F=1$ and $H=9$, but then adding $E$ and $G$ mustn't result in a carry, which is impossible, since $I$ can't be $8$ or $9$ and $E+G$ must be at least $3+5$.

So unless $A=1$, $(A,C)$ must be either $(2,3)$ or $(3,2)$. Also $B\notin\{1,5\}$. That leaves $10$ possible triples for $(A,B,C)$, which each determine values for $D$ and $E$, and it shouldn't be too much effort to go through those $10$ cases, find the $4$ remaining options for $F$, $G$, $H$ and $I$ and show that they don't work out.

That leaves us with $A=1$. In that case the options for $(C,B)$ can be enumerated as $(2,7)$, $(2,8)$, $(2,9)$, $(3,4)$, $(3,6)$, $(3,8)$, $(3,9)$, $(4,3)$, $(4,7)$, $(4,8)$, $(4,9)$, $(6,3)$, $(6,7)$, $(6,9)$, $(7,2)$, $(7,4)$, $(7,6)$, $(7,8)$, $(7,9)$, $(8,2)$, $(8,3)$, $(8,4)$, $(8,7)$, $(8,9)$. That's $24$ more cases, for a total of $34$ cases to be checked.

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  • $\begingroup$ @GiovanniResta: I think that comment should go under the question, not under this answer. $\endgroup$ – joriki Feb 22 '16 at 9:21

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