Number Puzzle

Originally posted by royalchicken
goldfish1, can you work out the mathematical basis of the trick? (Actually, you most likely can, but will you...?)

I 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.

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.

Originally posted by sarathian
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.

Following 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?