While surfing on the internet I found a divisibility rule for 7
Subtract 14 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 47, the original number is also divisible by 47. This too is difficult to operate for people who are not comfortable with table of 14.
I was thinking if we could prove it .Any help is appreciated
Answer
As $47\cdot3-140=1$
$$47\cdot3a-14(10a+b)=a-14b$$
$$47|(10a+b)\iff47\mid(a-14b)$$
See also : Divisibility criteria for $7,11,13,17,19$
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