Divisiblity by 7

[asifahmad]

First of all I would like to give a special thaks to SIR Malyy Goode (email address edited out) who has brought me to keep interaction with u[ the big and savant people ].i m really feel ecstatic.


actually this puzzle is poposed by me which published in one of the indian news paper TIMES OF INDIA (sUNDAY TIMES 12th sep
2004 ),and at that time i couldnt conclude myself that whether it is holding true for all sets of no or not ;but later it
is due to yrs great interest this elusive thing which i didnt pick up at the very bigining , so i m overwhelmed by u people that u have payed lots of attention
to make it rectified .thak u all .
but the fact is that unless any thing which is being framed is not confirmed
by others we cant make sure that it is true so the things r going in this manner ,and lastly i came up to catch the faults .
yr vailueable suggestions are whole heartedly accepted

it is very unfortunate that it doesnot hold good for no which r of six digits or multiple of six digit numbers since the A
system which works behind this conundrum is periodic of six digits, so the same thing is being repiteted

So the question should be
Let X is an integer of n digits.
Case 1) when n is not multiple of 6. Than the Integer
we get after repeating it 6 times is divisible by 7.
Case 2) when n is multiple of 6 and X is divisible by 7.
Than the Integer we get after repeating it 6 times
is divisible by 7.
Case 3) when n is multiple of 6 and X is not divisible by 7.
Than the Integer we get after repeating it 6 times
is not divisible by 7.

now a days i have been working for the programme development in his regard ,many things have completed but many r yet to established let us see what happes .
i m specially bothering abt six or its multiple digit numbers
yours comments are most welcome
asif ahmad
ISM DHANBAD
INDIA
3rd yr B.Tech
Mechanical Engg.

Originally Posted by akruppa

Oops, I'm wrong! The subgroups generated by 2 and 4 do not include -1, as the order of 2 and 4 is 3. This subgroup does, however, consist of {1,2,4}, which again sums to zero (mod 7).

a*x^5+...+a = a*(x^5+...+x^0). If a == 0 (mod 7), the result is trivially divisible by 7. Otherwise, the results is divisible by 7 iff x^5+...+x^0 is. ((Z/Zp)* is free of zero divisors, so a*b == 0 <=> a==0 or b==0).

If the order of x (mod 7) is even, the subgroup generated by x includes x^(ord(x)/2) == -1. As we are going through the different x^k, k=0, ..., 5, we generate
x^0 == 1, x^1 = x, ..., x^(ord(x)/2-1), x^(ord(x)/2) == -1, x^(ord(x)/2+1) == -x, ..., x^(ord(x)-1) == -x^(ord(x)/2-1)
Each element so generated can be paired up with it's negative, so they sum to zero.

If the order of x (mod 7) is odd, the case ord(x)=3 just happens to generate elements which again sum to a multiple of 7. The case ord(x)=1, i.e. x=7*k+1, does not generate multiples of 7 if a != 0 (mod 7): x^k == 1 for all k, so x^0 + ... + x^5 == 6 (mod 7), and the results will be 6*a (mod 7).

I.e. 10^6 == 1 (mod 7), and x=10^6 a=1 yields
1000001000001000001000001000001 = 3 * 19 * 101 * 9901 * 52579 * 333667 * 999999000001
No factor 7 here.

Alex