...There was another answer here that seemed close.

I'm not sure the best way to express it, but here's what I've got:

Allow me to use a shorthand, N = AB where AB is shorthand for 10 * A + B, and B is the least significant digit. IF N is divisible by 7, then AB = 0 (mod 7). This allows us to do the following manipulations:

  AB = 0 (mod 7)
  10*A + B = 0 (mod 7)
  3*A + B = 0 (mod 7)
  5*(3*A + B) = 5*0 (mod 7) // EDIT: had -2 on the RHS, that was correct but confusing
  A + 5*B = 0 (mod 7)
  A - 2*B = 0 (mod 7)

So we've arrived at drfuchs' method. You only need to repeat the process until you've proven to yourself that A - 2 * B is divisible by 7 or not.

That's all I've got. I ran a half-marathon yesterday, and I've been up for 16 hours today. My brain might work better tomorrow.

EDIT: changed the steps a bit, had some unneeded ones. Also, this shows that you can use addition instead of subtraction, but you won't be moving towards 0, but you still have to be able to recognize that the new number is divisible by 7.

50 years from now this will be called "drfuchs last theorem"

You are subtracting enough 21s (ie: X * 21) to make the last digit zero. Since 21 is obviously a factor of 7, subtracting those should yield a factor of 7.

You can then divide by zero because if X is a factor of 7 and ends in zero, X / 10 is a factor of 7 as well.

21=3*7

you are subtracting multiples of 21, then dividing by 10, which is coprime with 7