This is the first - presumably simplest - puzzle of an intended series of numerical dissection puzzles.

The goal is to dissect the 15 x 11 number-grid below into exactly 17 rectangular sub-grids.

  • Each rectangle has to be of shape of either ( ? x 3 ) or ( 3 x ? ) number squares.
  • Each rectangles has to fulfil either a sum-equation or a difference-equation
    using its 3 rows or 3 columns, respectively.
  • Numbers are either read "left-to-right" or "top-to-bottom" and never "right-to-left" or "bottom-to-top".
  • Leading zeros are acceptable.

Please also describe in your answer any "technique" or "routine" you have used to find the solution.


Examples of valid 'rectangles':

enter image description here

  • $\begingroup$ Nice puzzle, but I don't think I have enough spare time to try and solve it! That's going to be a long job for someone. $\endgroup$ May 20, 2015 at 8:41
  • $\begingroup$ @randal'thor actually, I don't think so. I believe there are several deductions one can make to systematically approach it. We will see. $\endgroup$
    – BmyGuest
    May 20, 2015 at 9:03

1 Answer 1


Wasn't all that difficult:

Solution to puzzle

To solve it, I started by marking out boundaries that I knew the solution would contain. For example, look at these numbers in the top left corner:

9 1 3 8 2 6 6 5 9 8 …
0 0 4 1 1 7 4 4 9 0 …
9 0 9 7 0 9 2 1 0 2 …
6 2 8 0 8 9 3 5 8 0 …
6 3 9 6 9 5 3 4 7 1 …
: : : : : : : : : :

Going from left to right, there is no way that we can form a valid sum of the form $9\cdots \pm 1\cdots = 3\cdots$, so we know immediately that the sum in the top left corner must be working from top to bottom; i.e., $913\cdots \pm 004\cdots = 909\cdots$. It turns out that we can carry this on for up to 9 digits, but we know for certain that there is a horizontal line below the digits $9\ 0\ 9$ at the start of the third row. Likewise, the two sets of three digits at the start of the fourth and fifth rows must form a horizontal sum, because $6\cdots \pm 6\cdots \neq 3\cdots$, but $6\cdots + 2\cdots = 8\cdots$.

Once you've identified a few boundaries this way, the rest fall into place quite easily.

  • $\begingroup$ I thought so (that it would be rather easy.) Funny enough, it is not exactly my solution, because I made a mistake. The top 9x3 block I have split into a 3x3 and a 6x3 but of course one can combine. On the other hand, you have two separate 3x3 in the bottom right on top of each other, which were a single 3x6 in my book, so the total of 17 is found again :c) Therefore a total dissection of only 16 blocks would be possible. I should have asked for the minimum I guess... But then again: I do these puzzles to learn about building them. Thanks for putting your time into it. $\endgroup$
    – BmyGuest
    May 20, 2015 at 10:47
  • 1
    $\begingroup$ The "subtraction" rectangles can be equivalently expressed as an "addition" rectangle with the arrow pointing in the opposite direction, making it easier to spot adjacent blocks of the same type... The solution posted has 2 sets of adjacent blocks of the same type, which can be combined (006305 + 717171 = 723476) and (2507200 + 3302820 = 5810030), allowing dissection into only 15 blocks. $\endgroup$
    – Steve
    May 20, 2015 at 10:53
  • 1
    $\begingroup$ @BmyGuest And thanks for posting it. It's a really neat idea for a puzzle. $\endgroup$
    – r3mainer
    May 20, 2015 at 10:59

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