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Topic: The divisibility rules of prime numbers
Replies: 2   Last Post: Apr 3, 2013 4:42 AM

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 shyamal kumar das Posts: 54 From: kolkata west bengal india Registered: 7/21/10
Re: The divisibility rules of prime numbers
Posted: Apr 3, 2013 4:42 AM

Dear Grei,Your letter post on Dec1,2012 escaped my notice. So, I could not reply in time. However, having read your comments, I believe that you have not gone through but glance through my article or, after reading first two pages, presumed that nothing is NEW.
Given below, is an annex to my original write-up. Hope, this may interest you to read my article once again. ANNEXE
TO THE DIVISIBILITY RULES OF PRIME NUMBERS
I have tried a different approach to find out if a given number is divisible by a prime factor.
Let the given number be: 10000a+1000b+100c+10d+e and
(A) the divisor is 11, which means n = -1
1st step: 1000a+100b+10c+d+(e*-1) = 1000a+100b+10c+d-e
2nd step: 100a+10b+c+(d-e)*-1 = 100a+10b+c-d+e
3rd step:10a+b+(c-d+e)*-1= 10a+b-c+d-e
4th step: a+(b-c+d-e)*-1 = a-b+c-d+e
(B) the divisor is 7, means n= -2
1st step: 1000a+100b+10c+d+(e*-2) = 1000a+100b+10c+d-2e
2nd step: 100a+10b+c+(d-2e)*-2= 100a+10b+c-2d+4e
3rd step: 10a+b+(c-2d+4e)*-2 = 10a+b-2c+4d-8e
4th step: a+(b-2c+4d-8e)*-2 =a-2b+4c-8 d+16e
(C) the divisor is 19, means n= 2
1st step: 1000a+100b+10c+d+(e*2) = 1000a+100b+10c+d+2e
2nd step: 100a+10b+c+(d+2e)*2= 100a+10b+c+2d+4e
3rd step: 10a+b+(c+2d+4e)*2 = 10a+b+2c+4d+8e
4th step: a+(b+2c+4d+8e)*2 =a+2b+4+-8 d+16e
(D) the divisor is 31, means n= -3
1st step: 1000a+100b+10c+d+(e*-3) =1000a+100b+10c+d-3e
2nd step: 100a+10b+c+(d-3e)*-3 = 100a+10b+c-3d+9e
3rd step:10a+b+(c-3d+9e)*-3= 10a+b-3c+9d-27e
4th step: a+(b-3c+9d-27e)*-3 = a-3b+9c-27d+81e
(E) the divisor is 29, means n=3
1st step: 1000a+100b+10c+d+(e*3) = 1000a+100b+10c+d+3e
2nd step: 100a+10b+c+(d+3e)*3= 100a+10b+c+3d+9e
3rd step: 10a+b+(c+3d+9e)*3 = 10a+b+3c+9d+27e
4th step: a+(b+3c+9d+27e)*2 =a+3b+9c+27 d+81e
Hence, it has been calculated and found that if the given number is 10^p a+10^p-1 b +10^p-2 c+10^p-3 d+???? , the simplified expression for checking the divisibility is:
1)a+nb+n^2 c+n^3 d+n^4e+???..(where n to be added)
or
2) a-nb+n^2 c-n^3 d+n^4e+???..(where n to be subtracted)
If the integral sum of the above expression is divisible by the Divisor, the given number is also divisible by the Divisor.
The process of simplification can be made easy to get the result. Here are the illustrations:

Example 1: To see if 11 is a divisor of 153967 ?
Solution: Here n= -1. All the integers of the given number is written column-wise and each one is multiplied by 1.
+ -
1*1= 1
5*1= 5
3*1= 3
9*1= 9
6*1= 6
7*1= 7
Total= 10 21
The integral sum=+10-21=-11, which is divisible by 11
Hence 153967 is divisible by 11
Example 2: To check If 97979 is divisible by 7?
solution: Here n= -2. All the integers of the given number is written
column-wise and each one is multiplied by 2,4,8,16 .
(2)9*1= 2
(0)7*2= 0
(2)9*4= 8
(0)7*8= 0
(2)9* 16= 32
----- ----
Total= 42 0
The integral sum= 42 ? 0 = 42 which is divisible 7. hence 97979
is divisible by 7.
The figures in the brackets are the nearby numbers to 7
or multiple of 7.
Explanation: If the given number is divisible by 7, the integral sum
is divisible by 7 and also, the numbers in brackets give
same result.

Example3: To verify that 4748359 is divisible by 7.
Solution: Here n = -2
4*1 = 4
7*2 = 14
4* 4 = 16
8*8(1)= 8
3*16(2)= 6
5*32(4)= 20
9*64(1)= 9
------ ------
Total= 35 42
The integral sum = 35 ? 42 = -7, which is divisible by 7. Hence
4748359 is divisible by 7.

Example: Show that 19 is a divisor of 265943?
Solution: Here n = 2
2*1= 2
6*2= 12
5*4= 20
9*8= 72
4*16(-3)= -12
3*32(-6)= -18
-------
Total= 76
The integral sum = 76, which is divisible by 19, hence
265943 is divisible by 19.
I am a teacher for last 48 years and I know what is taught in school about divisibility of numbers.

Date Subject Author
11/16/12 shyamal kumar das
12/1/12 grei
4/3/13 shyamal kumar das