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 2016-08-18, 12:03 #1 henryzz Just call me Henry     "David" Sep 2007 Cambridge (GMT/BST) 10110111011102 Posts Crossnumber I am currently solving this crossnumber and finding it very difficult. I also thought that this puzzle would interest many on the forum. https://issuu.com/chalkdust/docs/chalkdust-issue-03/47
2016-08-18, 13:03   #2
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by henryzz I am currently solving this crossnumber and finding it very difficult. I also thought that this puzzle would interest many on the forum. https://issuu.com/chalkdust/docs/chalkdust-issue-03/47
43D is 18
44A gives a hint that 29D ends with an even digit also since 44A and 29D meet at the start of 44A ( the last part of 29D) and 44A is half of 29D we also know that that even digit is half of the start digit of 29D so we are looking for multiples of 7 that are even starting with either 4 or 8 and ending with either 2 or 4 respectively. which makes 44A end with 1 or 2 respectively.

 2016-08-18, 13:35 #3 xilman Bamboozled!     "πΊππ·π·π­" May 2003 Down not across 2·72·109 Posts The old ones arre the best. Because seven ate nine.
 2016-08-18, 13:40 #4 henryzz Just call me Henry     "David" Sep 2007 Cambridge (GMT/BST) 2×5×587 Posts Have solved 11D, 15A, 23D, 30A, 34D, 39A, 42A, 43D so far, Down to two possibilities for 31D. 39D is bugging me
2016-08-18, 14:19   #5
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by henryzz 39D is bugging me
okay well it's 2 digits and doesn't have a sum to add to 1

let's see we can eliminate square numbers because they are n*n and n/n = 1 that eliminates 6 numbers out of 90. and 11 is taken out because it's 1 less than double a perfect number ( a number that can be defined as a number for which the sum of the reciprocal of the divisors sums to 2, the reciprocal of 1 is 1) so that also eliminates 55. so we have now taken out 8 numbers out of 90. and since xilman gave you 39 A we can cut to the chase and know that it starts with a 7. secondary edit: and the clue for 24 A shows us the sum of digits will be a two digit number and that doesn't work for secondary digits of 0,1,2, so the second digit must be greater than 2.

Last fiddled with by science_man_88 on 2016-08-18 at 14:47

2016-08-19, 22:55   #6
petrw1
1976 Toyota Corona years forever!

"Wayne"
Nov 2006
Saskatchewan, Canada

2×7×331 Posts

Quote:
 Originally Posted by henryzz Have solved 11D, 15A, 23D, 30A, 34D, 39A, 42A, 43D so far, Down to two possibilities for 31D. 39D is bugging me
Seems the answer is 23. But that doesn't fit with 39A (could be it anything other than 789?)
I can find a solution for every 7x number. Seems 23 is the highest 2 digit number though.

2016-08-19, 22:58   #7
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by petrw1 Seems the answer is 23. But that doesn't fit with 39A (could be it anything other than 789?) I can find a solution for every 7x number. Seems 23 is the highest 2 digit number though.
could be 678 in theory I guess or 786 or 78 <digit> or <digit>78 in which case 23 could work I'm trying to figure it out with PARI ( yeah I know I'm cheating).

Last fiddled with by science_man_88 on 2016-08-19 at 23:06

2016-08-19, 23:07   #8
Anonuser

Sep 2014

29 Posts

Quote:
 Originally Posted by petrw1 Seems the answer is 23. But that doesn't fit with 39A (could be it anything other than 789?) I can find a solution for every 7x number. Seems 23 is the highest 2 digit number though.
The following link leads to a clarification about 39D: http://chalkdustmagazine.com/categor...s/crossnumber/

By the way, I have attached a pdf-file which should contain the solution of the puzzle.
Attached Files
 solution.pdf (16.0 KB, 83 views)

Last fiddled with by Anonuser on 2016-08-19 at 23:22

2016-08-19, 23:34   #9
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by science_man_88 could be 678 in theory I guess or 786 or 78 or 78 in which case 23 could work I'm trying to figure it out with PARI ( yeah I know I'm cheating).
doh I forgot about the sum of digits clue so no 23 would never work.

Quote:
 19 28 29 37 38 39 46 47 48 49 55 56 57 58 59 64 65 66 67 68 69 73 74 75 76 77 78 79 82 83 84 85 86 87 88 89 91 92 93 94 95 96 97 98 99
are the only numbers that would work but since the solution has been given the point is nullified.

2016-08-21, 13:55   #10
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT/BST)

2·5·587 Posts

Quote:
 Originally Posted by science_man_88 are the only numbers that would work but since the solution has been given the point is nullified.
The solution is useful for checking results but we shouldn't use it to complete the puzzle.
I made a mistake. Once I have fixed that I will post my full progress in solving the puzzle.
There are some interesting programming challenges in this puzzle.

2016-08-21, 15:13   #11
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by henryzz The solution is useful for checking results but we shouldn't use it to complete the puzzle. I made a mistake. Once I have fixed that I will post my full progress in solving the puzzle. There are some interesting programming challenges in this puzzle.
sadly I can't help you much in one sense I made my own test to see what I knew in pari ( by asking yes or no to a question like do you know about or do you know how to use ... though I was wishy-washy on some based my answers I know how to use to some extent 184/832 PARI/gp functions so under 23% is what I know.

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