Monday, 22 September 2014

number theory - Proof for divisibility by 7



One very classic story about divisibility is something like this.





A number is divisible by 2n if the last n-digit of the number is divisible by 2n.
A number is divisible by 3 (resp., by 9) if the sum of its digit is divisible by 3 (resp., by 9).
A number ¯a1a2an is divisible by 7 if ¯a1a2an12×an is divisible by 7 too.




The first two statements are very well known and quite easy to prove. However I could not find the way on proving the third statement.



PS: ¯a1a2an means the digits of the number itself, not to be confused with multiplication of number.


Answer




5(\overline{a_1a_2\ldots a_n})=50(\overline{a_1a_2\ldots a_{n-1}})+5a_n=\overline{a_1a_2\ldots a_{n-1}}-2a_n\pmod{7}


No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...