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Old 2018-11-08, 18:49   #1
Till
 
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Default crossnumber #8

A challenging puzzle I found mentioned on https://blogs.ams.org/blogonmathblogs/ :

http://chalkdustmagazine.com/regular...mber-issue-08/

Last fiddled with by Till on 2018-11-08 at 18:52 Reason: reference
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Old 2018-11-08, 21:16   #2
science_man_88
 
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Quote:
Originally Posted by Till View Post
a few things.
9D,36A,20D,3D,4D,22D,8D,26D,23A,5D, and 6D have to all be 2 mod 5 and 6 mod 7 ( so 27 mod 35)
21A is 257
16D and 3A both show 18A<72, if different 18A<69
1A is surrounded by numbers that only have 2's and 3's as factors because n^2-1=(n-1)*(n+1)
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Old 2018-11-08, 22:42   #3
petrw1
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Quote:
Originally Posted by science_man_88 View Post
a few things.
9D,36A,20D,3D,4D,22D,8D,26D,23A,5D, and 6D have to all be 2 mod 5 and 6 mod 7 ( so 27 mod 35)
21A is 257
16D and 3A both show 18A<72, if different 18A<69
1A is surrounded by numbers that only have 2's and 3's as factors because n^2-1=(n-1)*(n+1)
I found this one MUCH easier than one I did a few months ago....only took me a few hours off and on.

Now lets hope I submitted the right sum

2 hints:
34A; Proth Numbers
14D: Triperfect number

Last fiddled with by petrw1 on 2018-11-08 at 22:51
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