Monday, 5 January 2015

elementary number theory - 11 divisibility



We know that 1331 is divisible by 11.
As per the 11 divisibility test, we can say 1331 is divisible by 11. However we cannot get any quotient. If we subtract each unit digit in the following way, we can see the quotient when 1331 is divided by 11.



13311331=132
132132=11
1111=0, which is divisible by 11.



Also the quotient is, arrange, all the subtracted unit digits (in bold and italic) from bottom to top, we get 121. Which is quotient.
I want to know how this method is working? Please write a proof.



Answer



Look how the method works for the number abcde. You substract e from abcd, getting abcd, and you apply the method to abcd. If the method works for abcd, it yields n such that abcd=11×n. But abcd0 is abcd×10=11×10×n and abcde=abcd×10+e×11 hence abcde=11×10×n+11×e is 11×(10×n+e). This proves that n=10×n+e is indeed the correct answer for abcde.



Edit (Upon OP's request, the same proof, with more apparatus but with zero more mathematics.)



Look how the method works for the number N=akak1a2a1a0 with k. You substract a_0 from a_ka_{k-1}\cdots a_2a_1, getting M=b_kb_{k-1}\cdots b_2b_1, and you apply the method to M. If the method works for M, it yields m such that M=11\times m. But b_kb_{k-1}\cdots b_2b_10 is M\times 10=11\times 10\times m and N=M\times 10+a_0\times 11 hence N=11\times 10\times m+11\times a_0 is 11\times(10\times m+a_0). This proves that n=10\times m+a_0 is indeed the correct answer for N if m was the correct answer for M. A recursion on the number of digits of N yields the result.


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