7
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enter image description here

This is a numeric crossword: all the answers are positive numbers with no leading zeroes. Each clue is given as a sequence of mathematical operations on letters, each of which represents a distinct prime number (e.g. if $A=2$ and $B=3$ then a clue might be written as $1.\ A^A+B\quad(1)$, to which the answer is $7$. Numbers in parentheses after the clue indicate the length of the answer, not necessarily the length of the grid entry.

The rules:

  • No grid entry starts with a leading zero
  • Before an answer can be entered into the grid the solver must perform a basic transformation on it
  • All answers and entries are unique
  • Once the solver has filled the grid they must choose a digit from all those appearing in the grid and shade all the cells corresponding to that digit. What appears must then be appropriately transformed and written below the grid in the space provided.
  • While parentheses have been provided for clarity, if there is any doubt then standard mathematical ordering of operations is meant (BODMAS)
  • Your lucky numbers for this puzzle are $30, 682$ and $15554$.

All feedback is welcome :)

Clues:

Across

1. $E\times (A+T) - (O-D)$ (5)

11. $D^S + D^D$ (3)

12. $((A\times H)-E)/O$ (2)

14. $(S\times E) - (N\times (A+T) + O)$ (4)

16. $H\times (A+D)$ (4)

18. $A^D - H +(D\times T)$(2)

19. $O^D \times T \times A -(O\times D)$ (3)

Down

1. $T^D$ (2)

2. $O \times S$ (2)

3. $D$ (1)

4. $N-O\times D\times D$ (2)

5. $N$ (2)

6. $D\times N - T\times O\times O$ (2)

7. $A-O$ (1)

8. $S$ (1)

9. $\sqrt{O\times N +S}$ (2)

10. $D\times (T^D - D\times A)^D$ (2)

13. $O^D$ (1)

15. $T$ (1)

17. $O$ (1)

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  • $\begingroup$ What is "a basic transformation" as given by rule 2? $\endgroup$ – RShields Jun 21 at 14:18
  • $\begingroup$ There's no way all variables can be one digit. 1-across is impossible if that's the case @OmegaKrypton $\endgroup$ – RShields Jun 21 at 14:24
  • $\begingroup$ oops... thanks @RShields $\endgroup$ – Omega Krypton Jun 21 at 14:26
  • $\begingroup$ Can you double-check 10 Down, 1 Across, and 19 Across? I'm getting a contradiction that probably involves at least one of those three. $\endgroup$ – Deusovi Jun 21 at 14:28
  • $\begingroup$ @RShields that's part of the puzzle :) $\endgroup$ – postmortes Jun 21 at 14:29
5
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First of all,

it's natural to guess that the "basic transformation" is, literally, a change of base, and a moment's consideration shows that if so the base must be 2. We can write in lots of 1s because no answer begins with a 0. Since $O^D$ is a single decimal digit O,D are 2,3 in some order. We have enough information also to deduce that S is 7 via 8d. Now the fact that 11a is 8 bits long tells us that D=2 and O=3 rather than the other way around, and we can fill in 11a. Since A-O is 10xx, A is in the range 11..16, so it's either 11 or 13. T is a 3-bit prime and isn't 7, so T=5. Now 6d is 2N-45 and its bits are 10xxx, and N is 11xx1; we quickly find that in fact N is 31. 19a is 45A-1 and the bits we know are enough to rule out A=13; so A=11. Now we have 1a = 16E-1, from which we can immediately fill in the last four bits (we already have two of them), and then the only way to fill in the blanks to make E prime is to make the other two be zeros and E=991. Now we can fill in a bunch of other things just by doing arithmetic. We end up with 18a being 1111x which requires x=0 and H=101 and can fill in everything else.

At this point the grid looks, if I haven't miscalculated, like this:

enter image description here

Now

the 1s in the grid appear to say SOS, which perhaps we should transform via a different kinda-binary code into ... --- .... postmortes, is everything OK? :-) A hint in the direction of this particular kinda-binary code comes from the "lucky numbers", which equal the products DOT, AND, DASH (using the letter assignments we have discovered in solving the puzzle).

Credit where due: Deusovi started on this before I did and worked faster than I did, and had he not run into errors in the original puzzle he'd doubtless have finished long before I did. (I haven't looked at anything in his answer, though, beyond checking whether the contradiction he found was the same as the one I did at a similar time.) Go and upvote his answer!

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  • $\begingroup$ Everything you say is right, except that you seem to have the calculation for 14A wrong. I've rechecked it in Excel and it's coming out right, so my guess is that you're adding O to (A+T) before multiplying by N? If I add extra brackets, you should be finding $(S\times E) - ((N \times (A+T)) + O)$ That gives you an odd number minus an odd number which is even, which fits your grid so far $\endgroup$ – postmortes Jun 21 at 15:43
  • $\begingroup$ Ah, no, actually an even simpler error -- I omitted the O. Thanks! $\endgroup$ – Gareth McCaughan Jun 21 at 16:01
  • $\begingroup$ Yes, that's all correct. If you factor the lucky numbers you'll get a hint for the encoding $\endgroup$ – postmortes Jun 21 at 16:07
  • $\begingroup$ Ah yes, so I do. Will edit accordingly. $\endgroup$ – Gareth McCaughan Jun 21 at 16:18
3
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Partial Answer (outdated by changes to the puzzle)

The trick to this puzzle is that

the "basic transformation" is binary: each answer must be converted to binary first before being entered.

The first step is to look at

the small numbers. From the short Down entries, T and S must be 5 and 7, and D and O must be 2 and 3.
From 1-Down, we need a two-digit power of either 5 or 7 that is 11XX1 in binary: this can only be 25 (from $5^2$), and so all four of those primes are determined.

Now that four of the variables are determined,

we look at 19-A. This must be $9*5*A$, where $A$ is a two-digit or larger prime. 11 is the only option that is small enough to fit in the grid, making 111101111. However, 10-Down must be even, and therefore must end in a 0, but the bottom right corner is a 1.

enter image description here

This seems to be a contradiction, but unfortunately I can't continue working on this at the moment.

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  • 1
    $\begingroup$ I interpreted the "basic transformation" the same way but found what seems a more elementary contradiction: 13d determines {O,D} though not which way around they are, which means neither of A,T can be a particular one of those, which constrains the end of 1a in an impossible way. But OP has fixed it now... $\endgroup$ – Gareth McCaughan Jun 21 at 14:49

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