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Subject: "Divisibilty Criteria for 7, 11, 13"     Previous Topic | Next Topic
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Thorsten Reimers (Guest)
guest
Dec-28-00, 07:14 PM (EST)
 
"Divisibilty Criteria for 7, 11, 13"
 
   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


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alexb
Charter Member
672 posts
Dec-28-00, 07:15 PM (EST)
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1. "RE: Divisibilty Criteria for 7, 11, 13"
In response to message #0
 
   Your divisibility criterion is incorrect. Try it for the number 1001 itself.


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Thorsten Reimers (Guest)
guest
Dec-28-00, 07:29 PM (EST)
 
2. "RE: Divisibilty Criteria for 7, 11, 13"
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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