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Forum URL: http://www.cut-the-knot.org/cgi-bin/dcforum/forumctk.cgi
Forum Name: Middle school
Topic ID: 33
Message ID: 2
#2, RE: Divisibilty Criteria for 7, 11, 13
Posted by Thorsten Reimers (Guest) on Dec-28-00 at 07:29 PM
In response to message #1
Hello,

you are right and I was wrong. You have to build the alternating sum of the groups

879 - 575 + 366 - 124 = 546

which is divisible by 7 and 11. The reason is that 1000 = -1 (mod 1001) So f.e. 575879 = 575 * 1000 + 879 = -575 + 879 (mod 1001).

Beg your pardon for my fault.

Best regards
Thorsten