# Problems With Divisors

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

• Could you please elaborate.I am not able to understand what do you mean? – S.Bansal Dec 29 '17 at 6:17
• I have written that a number is divisible by 47 if the last number(2-digit or 3-digit) at the end is divisible by 47 – S.Bansal Dec 29 '17 at 6:19
• After the process – S.Bansal Dec 29 '17 at 6:20
• A divisibility rule for 47, right? Also see here for a proof. – user371838 Dec 29 '17 at 6:20
• No problem, sir.. – user371838 Dec 29 '17 at 6:26

As $47\cdot3-140=1$
$$47\cdot3a-14(10a+b)=a-14b$$
$$47|(10a+b)\iff47\mid(a-14b)$$
• As $330-47\cdot7=1,$ $$33(10a+b)-329a=33b+a$$ – lab bhattacharjee Dec 29 '17 at 6:31