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 ¯a1a2…an is divisible by 7 if ¯a1a2…an−1−2×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: ¯a1a2…an 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}
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