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1. Sum of consecutive integer powers of 21 Across
7. Prime number, not all of whose digits are prime
8. Number that is coprime with 13 Down
9. Sum of three consecutive primes
10. Noble gas atomic number
12. 5 more than 9 Across
14. US coin denomination
15. Permutable prime
17. Fibonacci number
20. Power of 2 minus 1
21. Last two digits of 16 Down


1. 2 Down in base 21 Across
2. Leap year
3. Number not divisible by any of its digits
4. Number whose cube consists of three digits, each occurring twice
5. Palindromic number
6. 1 Across mod 7 Across
10. All 1-digit factors of 20 Across (excluding 1), in ascending order
11. 1 greater than 4 Down
13. Number that is less than 8 Across
18. Sum of digits of 8 Across
19. Can be expressed as sum of two squares, neither equal to 1, in two ways

Other notes:

  • Heavy lines indicate the beginnings and endings of entries.
  • No answers have leading zeros.
  • $\begingroup$ I was doing OK until I got to Rot13(gra qbja, gurer ner sbhe cbffvovyvgvrf sbe gjragl npebff ohg abar ba gurz jbex sbe gra qbja). $\endgroup$ – Weather Vane Oct 28 '20 at 21:14
  • $\begingroup$ @WeatherVane remind me how to decode that comment? $\endgroup$ – LarrySnyder610 Oct 28 '20 at 21:27
  • 1
    $\begingroup$ @LarrySnyder610 use rot13.com $\endgroup$ – bobble Oct 28 '20 at 21:33
  • $\begingroup$ @WeatherVane I think the clues and puzzle are correct. Note that gra qbja qbrf abg fnl cevzr snpgbef. $\endgroup$ – LarrySnyder610 Oct 28 '20 at 21:46
  • $\begingroup$ That's OK thanks, I can see what I did wrong now - but too late! Nice puzzle. $\endgroup$ – Weather Vane Oct 28 '20 at 21:47


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The first steps are clear.

16D: 315

21A: 15

17A: 317811

20A: 4095 (by 10D, this number is a multiple of 7)

10D: 3579

14A: 25

10A: 36

4D: 62 (by 11D, this number is between 59 and 68)

11D: 63

15A: 373

5D: 117711

13D: 134 (since it's smaller than 8A)

8A: 217 (by 18A, the sum of digits of this number is 10)

9A: 71 (by 12A, this number is between 65 and 74)

12A: 76

Here things become more tricky.

By 1D, an argument mod 25 tells us that the starting digit of 7A is 2 or 7.
Assume 7A starts with 2:

2D: 1612

1D: 727

6D: 176

At this point, I bruteforced the remaining two digits and saw that there was no solution.

Thus we conclude that

7A starts with 7, and hence

7A: 797

Now there are again two possibilities:

1D: 477 or 877.
Assume 1D: 477, then:

2D: 1012

7A: 709 (by 3D, the last digit cannot be 1)

Bruteforcing the remaining two digits again gives no solution.

Thus we get

1D: 877

2D: 1912

7A: 797

And finally

bruteforcing the remaining two digits gives the above answer (note that 6D cannot start with 0).

What I cannot verify:

1A: Sum of consecutive integer powers of 21 Across
I don't understand what this means...


I see, the sentence in 1A above is refering to the fact that $813616 = 15^0 + 15^1 + \dots + 15^5$. Thus my procedure above could be simplified a lot.

  • $\begingroup$ Correct solution, and very nice explanation. Your EDIT is correct, and my expectation was that that answer would come sooner in the progression, and that a lot of the other answers would follow from it. I didn't plan to require much or any brute force. Sorry if the wording in that clue was unclear! $\endgroup$ – LarrySnyder610 Oct 29 '20 at 12:16
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    $\begingroup$ @LarrySnyder610 Yes, once I understood the meaning, I realized that it should have required no bruteforce. I would suggest using the wording: repunit in base 21 Across. Even if one doesn't know the meaning of repunit, it is possible to search for it. $\endgroup$ – WhatsUp Oct 29 '20 at 14:28

Partial answer:

[4 down] 62 (= 238328^1/3)

[16 down] 315

[19 down] 85

[20 across] 4095 (= 2^12 - 1)

[21 across] 15

[17 across] 317811 (= 6-digit Fibonacci number ???8?1)

  • 2
    $\begingroup$ @bobble: my laptop crashed at that point and it took me until now to resurrect it and come back to finish it here - but why bother now. Downvotes not appreciated. I was about to post how the rest could be formulated as CSP and which items were Most-Constraining. I was the first to post any partial solution, and I did post six more solved items than you, or others. $\endgroup$ – smci Oct 28 '20 at 22:59
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    $\begingroup$ Small partial answers like this are generally discouraged -- see the meta post here for more details. $\endgroup$ – Deusovi Oct 29 '20 at 2:00
  • 1
    $\begingroup$ @Deusovi: like I already said, these were the first six clues solved here, I was in the process of solving it when my laptop started crashing, so this was all I had, if I hadn't posted it at the time it would have been lost. By the time I got my laptop resurrected (which took time) and came back to finish, someone else had already posted a solution, so I come back and see a downvotefest. As to the link you cite, you posted and immediately self-answered your own question, I wouldn't consider that generally representative, you should have waited a few days or week for other people's answers. $\endgroup$ – smci Oct 30 '20 at 2:20
  • 1
    $\begingroup$ And in this case, it certainly wasn't "low-effort rep-farming". IIRC it used to be standard practice on Puzzling to post substantial partial answers, right? (this one was as substantial as I could get before the laptop crashed) $\endgroup$ – smci Oct 30 '20 at 2:26

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