try this approach: 1000 = 142*7 + 6, so 1000 = (-1) mod 7.
Show that: A number N is congruent modulo 7 to the alternating sum of its digits in base 1000 (for example, 123456789 = 789 - 456 + 123 = 456 mod 7)
So... here's what I have (not much). Say the digits of N are dn ... d2 d1 d0, where each d is a digit that varies from 0 - 999 (using those as symbols basically because not enough letters).
Then N = (dn)(1000^n) +...+ (d2)(1000^2) + (d1)(1000^1) + (d0)(1000^0)
the end result is that i want this:
N = d0 - d1 + d2 - d3 + d4 -...+- dn mod 7.
Lets see if 7 divides this:
[(dn)(1000^n) +...+ (d2)(1000^2) + (d1)(1000^1) + (d0)(1000^0)] - (d0 - d1 + d2 - d3 + d4 -...+- dn)
which simplifies as:
d0(1000^0 - 1) + d1(1000^1 + 1) + d2(1000^2 - 1) +...+ dn(1000 +- 1)
But i don't know how to show that 7 divides this since dn could be anything...
Or maybe I'm doing this all wrong? Any suggestions?