I have a puzzle calendar that features 20 or so different types of puzzles. Some are pop culture references and some are logical. Anyway I can do most of the logical ones without breaking a sweat in under five minutes except for some subtraction puzzles. These really annoy me because there does not seem to be a logical way of doing the problem. I feel like you are forced to branch in your search space and essentially do an exhaust. After a nearly full year of frustration I realized that someone in this community may be just smarter at these than I am. So I'm posting one of the puzzles from my calendar. I'd like to know if there is a deductive method for solving these types of problems. I know the answer to this problem already (on the flip side of the paper) so don't bother posting that.

Puzzle: $$ \begin{array}{cc} & \square\square\square\square\square \\ - & \square\square\square\square\square\\ \hline = & 81975 \end{array} $$ Place the all the digits from $0$ to $9$ in the boxes, one time each, so that the subtraction is correct. Neither of the missing five-digit numbers may begin with $0$.

SOMETIMES I can see some way to make headway, like when the answer has a zero as a digit. But usually I can only see very painful ways to proceed that do not admit solutions with pencil and paper in five minutes or less (for me).


To make the explanation more convenient, let's rewrite the puzzle like this:$$ABCDE-FGHIJ=81975$$Consider the leftmost part: $A-F=8$. The possible answers are $9-1=8$ or $8-0=8$. But none of the numbers can starts with 0 so we take the former. Hence,$$9BCDE-1GHIJ=81975$$Now consider the fourth digit: $D-I=7$. You can think of $9-2=7$, $8-1=7$ and $7-0=7$. But $9$ and $1$ have been used already so we are only left with the last one. So,$$9BC7E-1GH0J=81975$$Consider the last digit: $E-J=5$. We now only have $2,3,4,5,6,8$ to choose from. Apparently, $8-3=5$. So,$$9BC78-1GH03=81975$$Consider the middle digit: $C-H=9$. Immediately we can think of $9-0=9$ but $9$ and $0$ are unavailable so we resort to things like $2-3=9$ and $5-6=9$ where the first number is one less than the second.
Consider the second digit : $B-G=1$. One can think of something like $6-5=1$ but bear in mind that we also have to take out one to supply the middle digit so indeed we want things like $6-4=1$ and $4-2=1$.
We now have $2,4,5,6$. If we choose $4$ and $6$ so that $6-4=1$, then we have $2-5\neq 9$ for the middle digit. So we can only arrange in the way such that $4-2=1$ and $5-6=9$. Thus, the final solution to the puzzle is $$94578-12603=81975$$

  • 3
    $\begingroup$ In the first two columns you work with, you ignore the possibility of borrows. In the tens column it could be $8-0$ with a borrow from the ones. It turns out not to happen here, but it does in other problems. The basic approach is good. $\endgroup$ – Ross Millikan Dec 24 '15 at 17:04
  • $\begingroup$ Yeah I give +1 for the effort, but it's the possibility of borrows that leads me to branch. $\endgroup$ – amcalde Dec 30 '15 at 15:06

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