Originally posted by royalchickenI believe , its mathematical basis lies in the remainder theorem of algebra. I am working on its details and will be back soon when something concrete comes out.
goldfish1, can you work out the mathematical basis of the trick? (Actually, you most likely can, but will you...?)
The mathematical basis of the " trick" (of divisibility by 7) is as follows.
Let (i) n be the number whose divisibility by 7 is to be tested.
(ii) a be its last digit.
(iii) n1 be the number left after truncating the last digit.
Now obviously n = 10 * n1 + a.
Further n1 - 2 a = 21*n1 - 20*n1 - 2 a = 21*n1 - 2 n.
Thus , if n1 - 2 a is divisible by 7, then 2n is divisible by 7 and hence
n is also divisible by 7.
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Originally posted by sarathianFollowing this method, one can ,similarly find similar tricks for divisibility by 13, 17, 19, .... etc. in exactly similar way . That will be quite useful and of practical benefit. Can you devise a similar trick for the binary system of numbers?
The mathematical basis of the " trick" (of divisibility by 7) is as follows.
Let (i) n be the number whose divisibility by 7 is to be tested.
(ii) a be its last digit.
(iii) n1 be the ...[text shortened]... divisible by 7 and hence
n is also divisible by 7.