Crossnumber1

[cinelli]

A crossnumber is like a crossword – just substitute numbers for words. The usual rule applies - no answer begins with a zero.

ACROSS (ac)                         DOWN (dn)
C. E dn - M ac                      A. A power of some number
H. Difference between B dn and      B. See H ac
   M dn                             D. A number possessing a common
I. K dn x 5                            factor with U ac
J. K dn / S dn                      E. C ac + M ac
K. (O ac)^3                         F. The sum of the cubes of two
M. E dn - C ac                         consecutive numbers
N. sqrt(N dn)                       G. The sum of the digits in Q dn
O. (K ac)^(1/3)                     I. A factor of P ac
P. A multiple of I dn               K. S dn x J ac
R. A palindromic number             L. U ac x 2
T. A dn + B dn + M dn               M. See H ac
U. L dn / 2                         N. (N ac)^2

Cinelli

[9873210]

There are no clues for O Q and S down. Is this a bug or a feature?

[9873210]

A down. A power of some number

x^1 = x, so this clue provides no information. The first digit of A is indeterminate.

[GoSeigen]

There are no clues for O Q and S down. Is this a bug or a feature?

S. K dn / J ac



GS

[UncleEbenezer]

A down. A power of some number

x^1 = x, so this clue provides no information. The first digit of A is indeterminate.

There are references to factors and multiples that could be similarly criticised. But A is also constrained by the clue for T.

Once upon a time I could do simple simultaneous equations in my head. Nowadays I think I'll have to write it down. So that'll mean digging up the back of an envelope ...

[cinelli]

There are no clues for O Q and S down. Is this a bug or a feature?

Sorry. This was my mistake. The full clues are

ACROSS (ac)                         DOWN (dn)
C. E dn - M ac                      A. A power of some number
H. Difference between B dn and      B. See H ac
   M dn                             D. A number possessing a common
I. K dn x 5                            factor with U ac
J. K dn / S dn                      E. C ac + M ac
K. (O ac)^3                         F. The sum of the cubes of two
M. E dn - C ac                         consecutive numbers
N. sqrt(N dn)                       G. The sum of the digits in Q dn
O. (K ac)^(1/3)                     I. A factor of P ac
P. A multiple of I dn               K. S dn x J ac
R. A palindromic number             L. U ac x 2
T. A dn + B dn + M dn               M. See H ac
U. L dn / 2                         N. (N ac)^2
                                    O. A multiple of U ac
                                    Q. See G dn
                                    S. K dn / J ac

Cinelli

[GoSeigen]

Spoiler:

First my rough notes showing my method -- sorry for any errors. Then the whole grid follows. I may have struggled more than necessary since I lacked one clue that Cinelli added later.

Notation:
Gd2 means Clue G down, second digit.
Gd means all of G down.


1. Gd2 = Ia5 = {0,5} so Gd ={10, 15, 20, 25}
2. 2010 < Kd < 10000 so Ka1 and Kd1 are not 1
3. Ld is at least 20 so Ka4 is not 1.
4. 12 < Oa < 20 because 3. prevents Oa=21, thus Oa1 = Od1 = 1
5. Ld= 2 Ua so is Ld is even and Na is even and Nd is a multiple of 4
6. From 5. and Ld = 2 x Ua ad is 2 digits, then Nd4 = Ua1 = {2,4}
7. Thus Na2 = Ld2 = {2,8} and Nd4 = Ua1 must be 4
8. 81< Ld < 99 so Ld1 = Ka4 = {8,9}
9. From 8, Oa = 19 is the only option so Ka = 19^3 = 6859
10. Ld = {92, 98} and Ua = {46, 49}
11. From 7. the only square of Na which gives the same initial digit for Nd so 98^2, so Na = 98 and Nd = 9604.
12. So Ld = 98 and Ua = 49; Ra2 = Ra5 = 0
13. From 9. Fd3 = 5 and the only sum of two consecutive numbers cubed ending in 5 and < 1000 is 8^3+9^3 = 855 = Fd
14. From 9., Ia1 = Id1 = 3
15. Thus Pa = {1,2,3} x Id and Pa4 = Qd1 = {9,8,7}
16. From 12. and 15. G must be 20 or 25.
17. 10 < Sd < 70 and 99 < J < 700
18. From Ta, Ta1 = Md4 = {1,2}
19. From Ha, Bd1 = Md1
20. Ia = 5 x Kd cannot be even because Kd4 would have to be {2,4,6,8} and no Kd4 ending in those digits will give Ia ending in 50. Therefore Ia5 must be 5 and Gd =25
21. Qd = {799, 889, 979}
22. From 17. and Ra, Kd4 = Ra3 = Ra4 =Sd1 = {1,3,5}
23. From 15 and 22, only Kd can end in {31, 33, 35, 61, 63, 65, 71, 73, 75, 91, 93, 95}; only 31, 71 or 91 can be multiplied by 5 to give Ia ending in 55. So Kd4 = 1 and Sd1 must also be 1.
24. Kd=Sd x Ja. Kd is odd so both Sd and Ja are odd. Kd4=1 so Ja and Sd must both end in 1 or 9, or one ends with 7 and the other 3. Sd = {11, 13, 17, 19}

XX25. Since Ja3 is 3 or more, then Pa = Id x 3 doesn't work as the total will be more than 10,000. So Pa must be a factor of one or two greater than Id.
25. A factor of two also doesn't work as no number ending in 9 (as Id) can be multiplied by 2 to give a number ending 68 (i.e. Pa).
26. Try a factor of 3. Then Pa1=9 and Pa4 = 7 and Q = 799. Ra = 901109
27. Pa ends in 67. Pa = Id x 3, so to get Pa ending 67 we need Ia to end with 89 (89*3=267). Id3 is 8. Id = [3089, 3189 or 3289}, other possibilities are too large.
28. If:
a. Id = 3089: Pa = 9267, then (from Ma) Ma4 = Ed5-Ca5 = 0 and Kd = 6091. But none of the Possible Sd is a factor of 6091.
b. Id = 3189: Pa = 9567, then (from Ma) Ma4 = Ed5-Ca5 = 3 and Kd = 6391. Sd=11 is a factor of 6091.
c. Id = 3289: Pa = 9867, then (from Ma) Ma4 = Ed5-Ca5 = 6 and Kd = 6691. None of the Possible Sd is a factor of 6091.


29. So it looks like Id = 3189, Pa =9567, Sd = 11 and Ja = 6391/11 = 581
30. A few steps follow quite easily: Ia = Kd *5 = 31955; Dd = {21, 91}; Ed = Ma + Ca so Ed = x9865.
31. Now try Md4 = 1. Then Ta5 = 1 = (5 + 1 + Bd4) mod 10 so Bd4 is 5. Hence calculate Ca3 = 2 giving Ed = 29865

32. Just discovered the bonus clues: the last part was getting really tough!

33. Od must be Ua x 4 = 196. Ha, that doesn't make it much easier!

34. Try Dd=91: now all possible values for Ca give out-of-range Ma.
35. Try Dd=21 (from 30): Then Ma = Ed - Ca = 29865 - 22282 gives Ma = 7583, all other possibilities for Ca leave Ma out of range. Ca = 22282 and Ma = 7583 are the only possibilities.
36. Try Bd > Md: Bd-Md gives Ha2 = 4 so Bd = x485.
37. From 19, Bd=7485.
38. Now Ta>16000 and is the sum of Bd, Md and Ad. Bd and Md are around 7500 so we'll try making Ad as large as possible. Since it's a power let's try 625 = 25^2, so H=24
39. We have Ta=625+7485+ 7461 < 16000, so this doesn't work.

40. So (ref 36.) try Md>Bd: Md-Bd gives Ha2 = 6 so Bd = x685.
41. From 19, Bd=7685.
42. Again, trying Ad=625, then Ha=26
43. We have Ta=625+7685+(7685+26) = 16021. This fits and completes the puzzle so I will take it as the solution.

Here is my complete grid:

6722282
2631955
5816859
7583698
7199567
1901109
1602149

For clue Pa I only tried a factor of three. A factor of one was another possibility, no time to see if it produced another solution.



Nice puzzle, goodness knows how it was generated but much more interesting than Sudoku or Wordle!

GS

[UncleEbenezer]

Working through this, starting with the more information-rich and closer-to-standalone clues,


Looking at F down, a three-digit sum of cubes of consecutive numbers must be 189, 341, 559 or 855.

Then we have K across, itself a four-digit cube. From the above, the third digit is 1, 5 or 9, so candidates are 2197 (13), 4096 (16), 4913 (17), or 6859 (19). So the first digit of L down is 3,6,7 or 9.

Now looking at the best clue: N across and down. The only candidates for a four-digit cube of a two-digit number sharing a first digit are 96/9216 or 98/9604. The clues for Ldn/Uac tell us Uac < 50, so it must be 98/9604. So Ldn > 2x40, and 98 is the only candidate, with Uac = 49. This also resolves Kac to 6859 and Fdn to 855.

With Oac = 19 and Odn a three-digit multiple of 49, Odn = 147 or 196.

Now we have one digit of Idn and of its multiple Pac. And similarly of Iac = 5*Kdn. We can deduce that the last digit of Kdn is 0 or 1. Since Rac is palindromic, that's also the first digit of Sdn, a 1. So Kdn = 6??1, and Iac = 3??55. And the first digit of Pac is odd.

Now we can use Qdn/Gdn to infer that the first two digits of the former add up to 6 or 16. The second digit is a 4 or a 9, so the first is 1 or 7. So for Pac we're looking for a multiple of 3??9 in the form ??61 or ??67. The multiplier can only be 3, so it's 9[258]67 and Idn is 3[012]89.

Now we have another digit of Kdn: it's 6?91. We also have it a multiple of Sdn, which is 1? Candidates are 11*581=6391, or 13*507=6591. The latter violates what we know about Idn, so it's the former. And Idn/Pac = 3189/9567, and Iac = 31955.

Ddn is a multiple of 7 ending in 1, so 21 or 91.

Adn ends in a 5, so it's 125, 225 or 625. Maybe we can use the 2 sometime, but not yet.

Now for a first bash at the really crap clues. Edn = Cac + Mac = ???82 + ??83. So we can fill Edn = ?9865. Mac being four digits, Ddn=91 yields no candidate subtraction, so Ddn = 21, and similarly the first digit of Mac is constrained to 7, and first digits of Cac and of Edn are equal (let's say X), and the second digit of Mac is 7 - X. Aaargh!

Finally we need to play with Bdn, Mdn, Tac et al. Bdn and Mdn differ by between 20 and 29. Adding the two of them to ?25, the second digit is a 6 and the final digit a 1. With the first digit of Mdn a 7, we can infer Tac = 2* 7??? + a three-digit number, so the first digit is a 1. And the final digits of Mdn and Bdn must add up to 6 (subtracting ?25 from 16??1), so the latter is a 5. Phew, now we have Mac = 7583, and from what we already knew, Edn and Cac are 29865 and 22282.

OK, Bdn = Mdn + Hac, knowing ??85 = 7??1 + 2?. So Hac = 24, Bdn = 7485, Mdn = 7461.

Finally apply the Tac clue again: 16??1 = 7485 + 7461 + ?25.

THis fails. OK, that's a sneaky clue at Hac, perhaps the wording can resolve to Mdn = Bdn + Hac?

Mdn = Bdn +Hac yields 7??1 = ??85 + 2? Then Hac = 26, Bdn = 7685, Mdn = 7711.
And then 16??1 = 7685 + 7711 + ?25. This time we have a solution. Adn > 605, so it's 625, and Tac is 16021. Phew!

In summary, from a cut&paste of the clues,

ACROSS (ac)                         DOWN (dn)
C. 22282                    A. 625
H. 26                            D. 21
I. 31955                           
J. 581                      E. 29865
K. 6859                      F. 855
M. 7583                        
N. 98                       G. 25
O. 19                     I. 3189
P. 9567               K. 6391
R. 901109             L. 98
T. 16021               M. 7461
U. 49                        N. 9604
                                    O. 196
                                    Q. 799
                                    S. 11

YOUR WIFE IS A BIG HIPPO!

[UncleEbenezer]

Looking at the previous solution, I find it instantly TL;DR. I daresay mine shares the same attribute to anyone who hasn't just posted it. I much prefer a simple and elegant solution, so if anyone can post one, ....

[GoSeigen]

Looking at the previous solution, I find it instantly TL;DR. I daresay mine shares the same attribute to anyone who hasn't just posted it. I much prefer a simple and elegant solution, so if anyone can post one, ....

I fear it's not that sort of puzzle. Very much like a standard crossword it's a matter of finding the easiest/most useful clues and then using the results to help with the others.

Uncle's criticism is justified as I just dumped my thoughts in a big list as I went along. Nevertheless his process is similar to mine -- the main difference being that I only discovered the clue for Odn when I'd almost completed the puzzle; knowing it earlier might have saved a few steps (and a lot of time!!!).


GS

[UncleEbenezer]

Uncle's criticism is justified as I just dumped my thoughts in a big list as I went along. Nevertheless his process is similar to mine -- the main difference being that I only discovered the clue for Odn when I'd almost completed the puzzle; knowing it earlier might have saved a few steps (and a lot of time!!!).


GS

I did much the same, with a bit of effort to sanitise what I posted, and go-back-and-delete a couple of times I'd misread the clues (eg confusing those for Iac and Idn - at which I almost cried foul that a multiple of 5 can't end in 9).

If I ever do another such puzzle, I need better logistics. I used the back of an envelope to fill in numbers and individual digits as I went along, and shifting between that and focus on the on-screen grid and clues left me rather stiff when I'd finished I suppose an option would've been to print out the original and the clues and do it all on paper

[cinelli]

Excellent solutions and reasoning from GoSeigen and UncleEbenezer. I admit I found this puzzle challenging and spent a few days over it (not full time, you understand). I have a couple more crossnumbers but haven’t managed to solve either yet. I am hesitating posting a puzzle I can’t solve myself.

Cinelli