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%I #3 Oct 01 2013 17:58:02
%S 111111,10101010101,1001001001001001,100010001000100010001,
%T 10000100001000010000100001,100000010000001000000100000010000001,
%U 10000000100000001000000010000000100000001
%N a(n) has the property that when multiplied by an appropriate n-digit number the product is the n-digit number repeated 6 times.
%C These numbers are divisible by 7 if the number of repeat digits d, is not a multiple of 6. Given the Sum = ((10^d)^6 - 1))/(10^d-1) from Fermat's little theorem, we note that (10^d)^(7-1) == 1 (mod 7) as required. The case when the repeat digits d = 6k we have the denominator of the Sum (10^k)^6 - 1 which by Fermat's little theorem again is divisible by 7. Thus 7 is canceled from the Sum and the sum is then not divisible by 7. It remains to prove this last statement. Also it appears 7 divides (10^k)^6 q+1 times when k = a*7^q. Obviously this method could be extended to divisibility by any prime.
%D The question arose on the mathforfun(AT)yahoogroups.com. Mansoor Anees asked: When an integer or a set of integers is repeated six times to form a new integer, it is divisible by 7. Why? For example, 121212121212 and 123123123123123123 and 444444 are divisible by 7.
%F Sum = (10^d)^5 + (10^d)^4 + . . . + (10^d)^1 + 1 = ((10^d)^6 - 1))/(10^d-1).
%e If n = 2 we have the number 10101010101. Multiply this by 13 to get 131313131313. and 131313131313/7 = 18759018759.
%o (PARI) divseven(n) = { for(x=1,n,y=((10^x)^6-1)/(10^x-1);if(y%7==0,print1(y","))) }
%K easy,nonn,base
%O 1,1
%A _Cino Hilliard_, Sep 13 2004
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