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Just a number crossword, but it has no Across clues, only Down clues. Leading 0s are allowed.

3 by 3 grid; top row has A, B, C, in the top left corners of their respective cells.

Down

  • A: Cube of the only digit not present in the puzzle
  • B: Product of four distinct primes
  • C: Important Well-known physics constant rounded to three digits.

Pretend I said "well-known" instead of "important" for clue C. It might be something you don't need to look up, even if you're not a physicist.

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  • $\begingroup$ A could be 343 or 512 $\endgroup$
    – The Empty String Photographer
    Commented Sep 2 at 19:38
  • $\begingroup$ @TheEmptyStringPhotographer Could it? Read carefully ;) $\endgroup$
    – AlephSquirrel
    Commented Sep 2 at 19:44
  • $\begingroup$ I just read that leading 0s are allowed, but that only means 027 is an additional possibility for A. $\endgroup$
    – The Empty String Photographer
    Commented Sep 2 at 19:52
  • $\begingroup$ Also, due to clue A, all digits in the crossword have to be unique $\endgroup$
    – The Empty String Photographer
    Commented Sep 2 at 19:55

2 Answers 2

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Given that there's only one digit that doesn't appear in the grid, there can be no repeats, so it was a quick task to go through all the cubes, and a slightly more tedious task to go through all the three-digit products of 4 distinct primes.

This is the grid I got

 0 5 9
 2 4 8
 7 6 1

With the answers obtained from

A: cube of 3, the only digit not in the grid
B: product of 2, 3, 7, and 13
C: One standard gravity in m/s^2, rounded to three digits

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    $\begingroup$ Ah, it would have been really useful if that constant had been in the list on the Wiki page I relied upon! The one you mention here should probably have sprung to mind on reflection... $\endgroup$
    – Stiv
    Commented Sep 2 at 20:43
  • $\begingroup$ Great minds think alike, but yours was faster or at least read the q earlier. +1. $\endgroup$
    – Oscar Lanzi
    Commented Sep 3 at 19:48
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The final 'crossword' should be filled like this:

5 6 3
1 9 7
2 0 4

The steps to fill it are actually quite tricky and require some logical case-bashing and trawling of data sources, like so...

Step 1:

We can pick up on a key rule straight away as a consequence of column 1's definition...

Since there is a solitary digit not present in the puzzle, each cell must be filled with a different digit 0-9, and one of the digits in that range will not be used at all. This means that none of our 3-digit answers can have a repeated digit.

This can be used to narrow down the candidates for A...

The cubes of the single digits are as follows (with reasons for exclusion if immediately obvious):

0 cubed = 000 NOPE (repeated 0)
1 cubed = 001 NOPE (repeated 0)
2 cubed = 008 NOPE (repeated 0)
3 cubed = 027 MAYBE
4 cubed = 064 NOPE (answer cannot include the digit that is its cube root, by definition)
5 cubed = 125 NOPE (ditto)
6 cubed = 216 NOPE (ditto)
7 cubed = 343 NOPE (repeated 0)
8 cubed = 512 MAYBE
9 cubed = 729 NOPE (includes digit that is its cube root)

This means we have two possible candidates for A: 027 (and so '3' is unused) or 512 (and so '8' is unused). Note that whichever of these we use, this number will make use of a '2' digit, so none of our other answers can use a '2'. Furthermore, our other answers cannot use 0 or 7 together with 1 or 5, since that would invalidate both established candidates for A. Keep all of these facts in mind (we'll need them)...

Step 2:

Next, consider B alongside a helpful table showing all the prime factorisations of 3-digit numbers. We are only interested in those that have exactly 4 distinct prime factors. By examining the table, the candidates are thus (with exclusions from the simplest applications of the established logical rules):

210 (2*3*5*7) NOPE (uses the 2)
330 (2*3*5*11) NOPE (two digits the same)
390 (2*3*5*13) MAYBE
462 (2*3*7*11) NOPE (uses the 2)
510 (2*3*5*17) NOPE (uses both 1 and 0)
546 (2*3*7*13) MAYBE
570 (2*3*5*19) NOPE (uses 0/7 and 5)
690 (2*3*5*23) MAYBE
714 (2*3*7*17) NOPE (uses 7 and 1)
770 (2*5*7*11) NOPE (two digits the same)
798 (2*3*7*19) MAYBE
858 (2*3*11*13) NOPE (two digits the same)
870 (2*3*5*29) MAYBE
910 (2*5*7*13) NOPE (uses both 1 and 0)
930 (2*3*5*31) MAYBE
966 (2*3*7*23) NOPE (two digits the same)

So we've narrowed it down to the following (with more complex exclusions to follow):

390 (2*3*5*13) – would force A=512 with no 8
546 (2*3*7*13) – would force A=027 with no 3
690 (2*3*5*23) – would force A=512 with no 8
798 (2*3*7*19) – would force A=512 with no 8 - CONTRADICTION, so NOPE
870 (2*3*5*29) – would force A=512 with no 8 - CONTRADICTION, so NOPE
930 (2*3*5*31) – would force A=512 with no 8

This leaves us with just 4 candidates - let's see what digits remain in these instances once we take out those we have used:

390 (2*3*5*13) – would force A=512 with no 8 i.e. ---4-67.--
546 (2*3*7*13) – would force A=027 with no 3 i.e. 1-.----89-
690 (2*3*5*23) – would force A=512 with no 8 i.e. --34--7.--
930 (2*3*5*31) – would force A=512 with no 8 i.e. ---4-67.--

This means the last column must contain the digits 189, 347, or 467 in some order.

Step 3:

To solve for C, let's consult a list of the main physical constants. Scanning the first 3 digits of all the entries in this list only produces one viable possibility...

The first radiation constant = 3.74 *10^-16

This fixes our answer for C and thus the others...

i.e. C must be 3.74, B must be 690 and A must be 512.

Or, to present the result as a grid:

5 6 3
1 9 7
2 0 4

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    $\begingroup$ A note to anyone reading this answer before Bass's - our final answers are entirely different, albeit having followed similar processes. While I like to think mine is still valid as the question is originally worded (re an 'important' physics constant), Bass's choice for C (unfortunately missing from the data source I consulted) is undeniably better known and so you should definitely go and read his answer too. $\endgroup$
    – Stiv
    Commented Sep 2 at 20:55
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    $\begingroup$ I would lean towards the gravitational constant as something you "don't" need to look "up", so Bass's answer is a little better to me. $\endgroup$
    – Oscar Lanzi
    Commented Sep 3 at 19:51
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    $\begingroup$ @OscarLanzi "don't" need to look "up", lol. There wasn't even supposed to be a pun there. Nice observation. $\endgroup$
    – AlephSquirrel
    Commented Sep 3 at 20:49

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