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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).

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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 $$94578-12603=81975$$

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    $\begingroup$ 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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