Two corollaries of ‘Divisibility of 11'

We notice that in decimal (10-base system), we can evaluate the divisibility of 9 by adding up all its digits and test the divisibility of the sum of the digits. While also in decimal system, we can evaluate the divisibility of 11 by adding up its odd digits and even digits and test the divisibility of the differences of the sum of the digits.

Following the logic in the article that discussing the divisibility of 9, we can made the following corollary: Similar to the idea that we can test test the divisibility of the differences of the sum of the digits, we could also test the divisibility of 12 by the differences of the sum of the odd and even digits in a base 11 system, the divisibility of 13 by the differences of the sum of the odd and even digits in a base 12 system…. etc.

Moreover, I made the hypothesis that in decimal system, divisibility by 12 could be tested by the divisibility of the weighted differences of the sum of the digits, i.e. even digits sum*2- odd digits sum. Moreover, divisibility by 13 could be tested by the divisibility of the weighted differences of the sum of the digits, i.e. even digits sum*3- odd digits sum, divisibility by 14 could be tested by the divisibility of the weighted differences of the sum of the digits, i.e. even digits sum*4- odd digits sum…. etc.
And in general, divisibility by n+x in a n-base system could be tested by the divisibility of the weighted differences of the sum of the digits, i.e. even digits sum*n- odd digits sum(in the same system.)

Modulus of 9

We all learn from our school that to test the divisibility of a number by 9, we just need to add up their digits and see if the sum of digits is divisible by 9. It is actually beyond that, the remain of a number divided by 9 is equal to the remain of the sum of digits by 9, which we can prove here:

Assume the digits are n(1)n(2)n(3)….n(k) in decimal system,
The value of number= n(1)*10^(k-1)+n(2)*10^(k-2)+n(3)*10^(k-3)+….+n(k)
= n(1)*[10^(k-1)-1]+n(2)*[10^(k-2)-1]+n(3)*[10^(k-3)-1}+....+[n(1)+n(2)+n(3)+....+n(k)]

Given that 10^n-1 is also divisible by 9 when n>=1,
thus the remain of the number divided by 9 is equal to the remain of the sum [n(1)+n(2)+n(3)+....+n(k)] divided by 9
which means the remain of the sum of its digits in decimal system when divided by 9 is equal to the remainder of the value of the number when divided by 9.

To give an example,
12345678=1*10^7+2*10^6+3*10^5+4*10^4+5*10^3+6*10^2+7*10^1+8
=(1*10^7+2*10^6+3*10^5+4*10^4+5*10^3+6*10^2+7*10^1+8 )
=((1*9999999+1)+(2*999999+2)+(3*99999+3)+(4*9999+4)+(5*999+5)+(6*99+6)+(7*9+7)+8 )
=(1*9999999+2*999999+3*99999+4*9999+5*999+6*99+7*9)+(1+2+3+4+5+6+7+8 )
(1*9999999+2*999999+3*99999+4*9999+5*999+6*99+7*9) is divisible by 9
thus 12345678 modulus 9=(1++2+3+4+5+6+7+8 ) modulus 9=0
12345678 is divisible by 9

There is nothing special about that, however, if we consider a number in k-base system,
i.e. The value of 12345678(k)=(1*k^7+2*k^6+3*k^5+4*k^4+5*k^3+6*k^2+7*k^1+8 )

Consider that 12345678(k)=(1*k^7+2*k^6+3*k^5+4*k^4+5*k^3+6*k^2+7*k^1+8 )
=[1*(k^7-1)+2*(k^6-1)+3*(k^5-1)+4*(k^4-1)+5*(k^3-1)+6*(k^2-1)+7*(k^1-1)]+(1+2+3+4+5+6+7+8 )

By the remainder theorem,
[1*(k^7-1)+2*(k^6-1)+3*(k^5-1)+4*(k^4-1)+5*(k^3-1)+6*(k^2-1)+7*(k^1-1)] is divisible by k-1,
i.e. when k=1 is substitute into the polynomail, it give the value of zero.

Thus the modulus of 12345678 in k-base system when divided by k-1 is equal to the modulus of (1+2+3+4+5+6+7+8 ) when divided by k-1

Therefore we can generalize from this result that in 10-base system, we can test the divisibility of a number by 9 by testing its divisibility of its sum of digits. We could also test the divisibility of a number in 9 base-system by 8 by testing its divisibility of its sum of digits when divided by 8. We could also test the divisibility of a number in 8 base-system by 7 by testing its divisibility of its sum of digits when divided by 6;…. etc.