enter image description here The grid above shows the final layout of a domino game played with a standard $28$-tile (dominos $0-0$ to $6-6$). The first domino played, the $4-4$ double is given, and solvers must complete the remainder of the grid. The red-shaded space is where the last two dominos in the set would need to go -- but sadly, though they would complete the layout there is not enough space for both of them. Solvers must also identify these two missing dominoes. To help solvers a little, doubles are the only dominoes that can change the direction of the layout and are set with their centre against the new direction. Looking at the start piece, heading left or right is a new direction, while moving up or down is in conformity with the layout.

As always, two dominoes may only abut if their numbers match.

To further assist solvers, crossnumber-style clues are provided below. They are split into Up, Down, Left and Right and indicate which direction the solver must eventually enter them into the grid. For example, if the solver were to find that $4113$ was an up-domino then the $4$ is entered in a cell, then the solver must move up to enter $1$, up to enter $1$ and up once more to enter the $3$. No entry goes round corners. Left dominoes are entered by moving left (West, if North is directly up the page), and Right dominoes are entered by moving right (East).

Solvers must perform a transformation on all answers before entering them into the grid, but this will become obvious as the clues are solved.

Notation: Capital letters represent positive integers with no leading zeros. Lower-case letters indicate clues: $2$u means $2$up, i.e. clue number $2$ from the up column. $\#$ indicates concatenation: $4\#1 == 41$. Numbers in brackets give the length of the answer (digit count) before transformation and grid entry.


$$ \begin{eqnarray} {} & \mbox{Up} & {} & {} & \mbox{Down} & {} & \\ 1. &\small \ (M \times N)\# (2u \times N) & (4) &\hspace{3cm} 1. &\small \frac{D}{O}+M &(2) &\\ 2. &\small (M \times O) - S & (2)&\hspace{3cm} 2. &\small \frac{M \times O}{N}& (2) &\\ 3. &\small \frac{D-S}{M} + N & (2)&\hspace{3cm} 3. &\small ((N^N)^N \times M)\# N & (4) &\\ 4. &\small O^N \times (2u-M) & (3)&\hspace{3cm} 4. &\small O \times M^N +M & (3) &\\ \\ {} & \mbox{Right} & {} & {} & \mbox{Left} & {} & \\ 1. &\small (3u) \# (D\times N^O) & (5) &\hspace{3cm} 1. &\small D+O &(3) &\\ 2. &\small (2d-S)\# N^O\# N^N & (4)&\hspace{3cm} 2. &\small (2u-M)\# \left((2u-M)\times \left(\frac{M}{O+N} + O\right)\right) & (4) &\\ 3. &\small N \# O \# O^N \# (O+N^N) & (4)&\hspace{3cm} 3. &\small (M+S)\times O & (2) &\\ 4. &\small \left(\frac{M}{N} \right)^N & (2)&\hspace{3cm} 4. &\small (2u-M)\# \left(D\times S \times \frac{M}{N}\right) & (5) &\\ \end{eqnarray} $$

No answer, before or after transformation, has leading zeroes, and division is always exact (there will never be $25/3$ for example, only $25/5$). Each capital letter represents a different positive integer.

Final note on tags: there's no cross-number tag so I've used cryptic-crosswords as numeric cryptic crosswords are traditionally like this.

  • $\begingroup$ A few clarification requests. 1. In e.g. 2L, can we specifically assume that $\frac M{O+N}$ is an integer and not merely that we get one when we multiply that by $(2u-M)$? 2. Concatenations are all base 10, right? $\endgroup$
    – Gareth McCaughan
    Commented Dec 6, 2019 at 19:14
  • $\begingroup$ @GarethMcCaughan Yes to both questions $\endgroup$
    – user40528
    Commented Dec 6, 2019 at 19:16
  • $\begingroup$ Further clarification request: when we fill in the grid using the solutions to the clues, we expect there to be a gap where that red bit is? $\endgroup$
    – Gareth McCaughan
    Commented Dec 6, 2019 at 19:35
  • $\begingroup$ Yes, you will be missing two dominoes from the set that should go there where there’s space for only one $\endgroup$
    – user40528
    Commented Dec 6, 2019 at 19:42
  • $\begingroup$ How do you feel about computer-assisted solutions? (For the moment I'm doing everything by hand other than isolated bits of arithmetic.) $\endgroup$
    – Gareth McCaughan
    Commented Dec 6, 2019 at 19:49

1 Answer 1


[I haven't spoilered any of what follows other than the images. It shouldn't be difficult to ignore a big wall of text like this.]

First of all, note that despite the "transformation" (which I expect often to change the lengths of our numbers) we should be able to rely on the given lengths, because we're told explicitly they they are the pre-transformation lengths.

If $N>2$ then 3d is at least $(3^3)^3$ which is well over 4 digits. So $N=1$ or $N=2$. On the other hand, if $N=1$ then 3d is just $M\#1$, requiring $M$ to be 3 digits, but considering e.g. 1d or 3l shows that this is impossible. So in fact $N=2$.

So 4r tells us that $10\leq(M/2)^2<100$ so $7\leq M<20$, and that $M/2$ is an integer; so $M$ is one of 8,10,12,14,16,18.

Now 3r tells us that $O$, $O^2$, $O+4$ are all single digits, so $1\leq O\leq3$. And 4u tells us that $O^2(2u-M)$ is 3 digits while $2u$ itself is known to be only 2 digits, so $O>1$. That is, $O$ is 2 or 3. And (remark: I hadn't noticed this when I started solving...) all our capital letters represent different numbers and we already have a 2, so $O=3$.

We know from 2l that $M$ is a multiple of $O+2$ as well as of 2. So in fact we must have $M=10$.

We know from 2r that $2d-S$ is 2 digits (the other concatenands being 44 or 84). And 2d is 15; since subtracting $S$ from this still yields a 2-digit number we have $S\leq5$.

From 2l it's clear that $2u-M$ is 2 digits long. So 4l tells us that $D\times S\times\frac MN\leq999$; we know that $\frac MN=5$ so this means that $DS\leq199$. And 1l tells us (given that $O=3$) that $D\geq97$, so $S\leq2$. 2 is already taken so $S=1$.

Now we know from 1d that $D$ is a multiple of 3, and from 3u that $D-1$ is a multiple of 10. That is, $D$ is 21 mod 30. From 1r we know that $8D$ is 3 digits long so $D\leq124$, and from 1l we know that $D\geq97$. The only value consistent with these is $D=111$.

At this point we have solved all the clues.
Up: 2058, 29, 13, 171.
Right: 13888, 1484, 2397, 25.
Down: 47, 15, 1602, 310.
Left: 114, 1995, 33, 19555.

Of course these have lots of out-of-range digits, so the obvious thing to do is to convert them to base 7. We get:
Up: 6000, 41, 16, 333.
Right: 55330, 4220, 6663, 34.
Down: 65, 21, 4446, 622.
Left: 222, 5550, 45, 111004.

Unless I am misunderstanding something, there is no difference between e.g. an Up and a Down number except for the direction of entry (and in particular there is no assumption that we can go around the whole loop in a consistent direction entering these numbers in the direction we're travelling), so what we have is equivalent to:

Down: 0006, 14, 61, 333, 65, 21, 4446, 622.
Right: 55330, 4220, 6663, 34, 222, 0555, 54, 400111.

Our set of dominoes has 8 of each number. The numbers above have, in total, 9,5,9,8,7,7,7 of numbers 0..6. Clearly we have some overlap (which isn't a surprise) and clearly even after removing our two extra dominoes there are at least a couple of digits we need to infer by other means.

The first thing we can place is 0006 (down). It must contain a 00; the only double that isn't at one end is the 44 in the middle (note that there are exactly 7 doubles and we can see where they all have to go), so this divides as 00|06 and goes at top right.

Next, 333 (down) must go at bottom left. 4446 goes in the middle; we can't have 4|6 because dominoes must match where they abut so it's 44|46. At this point we have all the vertical doubles we can have, which means that 622 (down) splits as 62|2. And of course 55330 (right) must split as 5|53|30 since we have a 33 elsewhere.

Where can 400111 (right) go? Not at centre-left where we have no space longer than 4. There's a double somewhere near its right end, which must in fact be at the right end, so this is centre right or bottom centre and divides as 40|01|11. Similarly, 6663 (right) must contain a double 6, which must be right at the end (66|63), and this must go at centre left or top centre. 0555 (right) must contain a double 5, which must be at the end, and this must go at centre right or bottom centre. 222 contains a double 2 and can go in whichever horizontal double slot remains (so of course 4220 divides as 42|20).

Since we have 42|20, the 02 (down) must in fact be split: 0|2.

Now, actually 40|01|11 can't go at centre right, because that would require a 4 to its left because of the abutting rule, and the other number on that domino would also have to be a 4 because of the abutting rule, and that would make the tile a second 44, which is impossible. So 40|01|11 goes at the bottom, and then 05|55 must go at centre right. We can deduce a couple of other dominoes using the abutting rule. At this point the picture is as follows (except that, duh, I've omitted one 0 we can deduce from the abutting rule):

enter image description here

Now, the only 4-dominoes still unplaced are 14, 24, and 45. They must go in those spaces in the middle. And we know that the latter actually appears in 42|20. There's only one way to place this: on the right. 14 and 54 will go in the other two spaces. Which is which? We know that 54 appears rightward, and it can't be across a domino boundary because of the abutting rule, so 54 must be in that space on the left and 14 in the nearby downward space.

The only remaining 0-domino is 03, so that goes up in the top right; and we know that that's part of 5|53|30, so we can place all of that.

enter image description here

The two remaining doubles are 22|2 and 66|63, both rightward. We can't put 66|63 at the top because it would conflict with the 53 to its right, so it goes at the left and 22|2 at the top. Now, after a bit of abutting, the only thing left unaccounted for is 62|2 (down), which goes in the only remaining space, on the left:

enter image description here

The gap we would like to fill wants a 3 on the left and a 5 on the right. The two remaining dominoes are 31 and 15. We could finish the job with just a little non-euclidean geometry :-).

  • $\begingroup$ Absolutely fantastic write up! The solution is correct, and I'm quite relieved to see that the grid fill went smoothly once all the clues were correct so the numbers were known. I hope you enjoyed it, after we got the corrections made! $\endgroup$
    – user40528
    Commented Dec 7, 2019 at 11:10
  • $\begingroup$ Yup, it was a fun solve. $\endgroup$
    – Gareth McCaughan
    Commented Dec 7, 2019 at 11:11

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