#0, Divisibilty Criteria for 7, 11, 13
Posted by Thorsten Reimers (Guest) on Dec-28-00 at 07:14 PM
Hello,I am missing the following, well known, divisibility criteria: If you have a number with many digits you can split them into groups of three digits and add these groups. Apply the method to the sum again if it is greater than one thousand. The original number is divisible by 7, 11 and 13 if and only if the final sum is. Example: Original number 124366575879 Groups of three digits 124 366 575 879 Sum is 1944 Final sum is 945 It is divisible by 7 and 13 but not by 11, so the same apply to the original number. The proof of this criteria is based on the fact that 1001 = 7·11·13. Best regards
Thorsten
#1, RE: Divisibilty Criteria for 7, 11, 13
Posted by alexb on Dec-28-00 at 07:15 PM
In response to message #0
Your divisibility criterion is incorrect. Try it for the number 1001 itself.
#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
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