We all hear of how test the divisibility of any number by 11 by adding its odd digits and even digits up, then if the difference between the sum of odd digits and the sum of even digits is a multiple of 11. We could therefore deduced that the original number is divisible by 11. Here we attempt to prove it as the consequence of decimal system as well as to notice a few other facts.
To prove this property, there are two ways to do this: We can either create a multiply of 11 then notice the numerical property of its digits, or we can prove backward from the numerical property of its digits to the number made up by such digits to be a multiply of 11. It doesn’t matter which way we does since one always implied another.
Consider the easier way, let’s make a multiple of 11, say 11*360
We have 11*360=3600+360=3960,
We notice because we are decimal system, and 11 indicate 10+1 or 1 exceed the base of 10, therefore the multiplication is simply moving the multiple number(360) one digit forward to become 3600, then added this number to it original number.
That is essentially what we all do in the multiplication by 11.
We notice that the sum of its odd digits=0+9=9
And the sum of its even digits=3+6=9
The difference=9-9=0 which is divisible by 11.
Assume a random number to be multiple by 11, say ABCDE in decimal system.
The process is 11*ABCDE=ABCDE0+ABCDE,
Take a look at the new digits of the result in descending order: A,A+B,B+C,C+D,D+E,E
We notice something curious, because it appears that the first digit of the number ABCDE appear twice in first and second place, the second digit of the number ABCDE appear twice in second and third place, then the third digit of the number ABCDE appear twice in third and fourth place; fourth digit of the number ABCDE appear twice in fourth and fifth place, and lastly fifth digit of the number ABCDE appear twice in fifth and sixth place. That is all of the consequence of multiplication of 11 in decimal system.
Seeing this pattern, to obtain a sum of all digits without repeating the digits from ABCDE is to adding the alternative digits, thus:
The sum of even digits=(A+B)+(B+C)+(D+E);
The sum of odd digits=A+(B+C)+(D+E)
The difference of them=(A+B+C+D+E)-(A+B+C+D+E)=0, which translated to the digits of the result is compose of digits repeating themselves. So multiplication of 11 would logically arrive in the difference of the sum of digits being 0, which is divisible by 11.
Notice that we assume so far that the sum of digits doesn’t exceed ten, therefore we can nicely summarize the digits of the resulting number as A,A+B,B+C,C+D,D+E,E. That is unlikely since some parts of the sum may exceed 10, therefore the digits of the resulting number could be A,A+B,B+C,C+D+1,D+E-10,E.
Actually, we don’t need to repeat the process of summing the odd and even digits of the resulting number to realize that the odd sum and even sum is differ by 1-(-10) or 11. And 11 is definitely a multiply of 11.
Assume we have more case like that, therefore the resulting digits is: A+1,A+B-10+1,B+C-10+1,C+D-10+1,D+E-10,E.
The sum of the odd digits=E+(C+D-10+1)+(A+B-10+1)=A+B+C+D+E-18;
The sum of the even digits=A+1+(B+C-10+1)+(D+E-10)=A+B+C+D+E-18
Therefore their difference is zero.
Clever reader may started to try the case of different digits of sum exceed ten, this seem like an dead end to me since when extend this result to a n-digits number, we had to try close to 2^n times. So what appear as an easy way turn out to be not so easy.
Suppose we started from the result of multiplication by 11. That give us a number with the difference of the sum of even and odd digits is the multiply of 11.
i.e. Given n(1)n(2)n(3)n(4)….n(2k),
(n(1)+n(3)+n(5)+….n(2k-1))-(n(2)+n(4)+n(6)+….n(2k))=11s (where s is a natural number)
we could write: (n(1)+n(3)+n(5)+….n(2k-1))=11s+(n(2)+n(4)+n(6)+….n(2k))
Suppose we let s(1),s(2),s(3) such that s(1)+s(2)+s(3)….+s(k)=11s,
so n(1)=n(2)+s(1), n(3)=n(4)+s(2), n(5)=n(6)+s(3)…., n(2k)=n(2k)+s(k)
The value of the original number equal to: n(1)*10^(2k-1)+n(2)*10^(2k-2)+n(3)*10^(2k-3)+….+n(2k)
Substitute back the n(odd) in terms of n(even), we have:
n(1)*10^(2k-1)+n(2)*10^(2k-2)+n(3)*10^(2k-3)+….+n(2k)= (n(2)+s(1))*10^(2k-1)+n(2)*10^(2k-2)+(n(4)+s(2))*10^(2k-3)+….+(n(2k)+s(k))*10+n(2k)
=((n(2)+s(1))*10+n(2))10^(2k-2)+((n(4)+s(2))*10+n(4))10^(2k-4)+….+(n(2k)*11+s(k)*10)
=(11*n(2)+10*s(1))*10^(2k-2)+(11*n(4)+10*s(2))*10^(2k-4)+(11*n(6)+10*s(3))*10^(2k-6)+….+(n(2k)*11+s(k)*10)
To test the divisibility of this number by 11, we can first remove all the multiple of 11 from value. Since when a number divisible by 11, and when take away the multiple of 11 from that number, the result would still be divisible by 11.
Therefore the value we can test is: (10*s(1))*10^(2k-2)+(10*s(2))*10^(2k-4)+(10*s(3))*10^(2k-6)+….+(s(k)*10)
=[s(1)*10^(2k-2)+s(2)*10^(2k-5)+s(3)*10^(2k-6)+....+s(k)]*10
={(s(1)+s(2)+s(3)+…+s(k))+[(s(1)*(10^(2k-2)-1)+(s(2)*(10^(2k-4)-1))+(s(3)*(10^(2k-6)-1)+....+99*s(k)]}*10
By the assumption the sum of s(1)+s(2)s(3)+…+s(k)=11s, and
We also notice that 10^(2m)-1 has even digits of 9, therefore it is divisible by 11.
Therefore [(s(1)*(10^(2k-2)-1)+(s(2)*(10^(2k-4)-1))+(s(3)*(10^(2k-6)-1)+....+99*s(k)] is the sum of multiplies of 11, it must itself divisible by 11.
Since the sum of multiplies of 11 and multiples of 11 must also be a multiplies of 11, therefore the value
n(1)*10^(2k-1)+n(2)*10^(2k-2)+n(3)*10^(2k-3)+….+n(2k) must be a multiply of 11. i.e. The number with digits n(1),n(2),n(3)….,n(2k) is divisible by 11.
Notice the above result only applied to number of even digits, to extend the applicability of the above proof. We only need to let n(1)=0, since the sum of odd digits and even digits would not changed by adding a zero term. Therefore the above result applied for both odd and even digits of number in decimal system.
A more complex method is to actually do the long division by 11 of the number with assumed properties of the digits. I would guess I left that for another day.
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