987654321	=	8	x	123456789	+	9
98765432	=	8	x	12345678	+	8
9876543	=	8	x	1234567	+	7
987654	=	8	x	123456	+	6
...
9	=	8	x	1	+	1

Saul Foresta explained as follows why this pattern holds in the decimal system and in other number systems using bases other than base 10:

I generalized the problem for any base n and any number of digits r where r can be anywhere from 1 to (n - 1).

Then after rewriting both sides of the equality given in the problem using sigma notation I arrived at the following:

\sum_{k = 1}^{r} (n - k) n^{r - k} = (n - 2) \sum_{k = 1}^{r} {kn}^{r - k} + r

In each summation k stands for the kth digit of the number we're dealing with, reading from left to right. For example, in the number 9876, k ranges from 1-4, where 9 is k=1, 8 is k=2, and so on.

So all I need to do in order to prove that this pattern holds is show that the left side of this equality does indeed equal the right side. Taking the terms like 8 x 123456789, that is

(n - 2) \sum_{k = 1}^{r} {kn}^{r - k}

over to the left hand side, we will prove that this expression is equal to r.

[(n-1)n ^r-1 + (n-2)n ^r-2 + (n-3)n ^r-3 + ... + (n-r)] - (n-2)[n ^r-1 + 2n ^r-2 + 3n ^r-3 + ... + r] =
[n ^r - n ^r-1 + n ^r-1 - 2n ^r-2 + n ^r-2 - 3n ^r-3 + ...+ n - r] - [n ^r - 2n ^r-1 + 2n ^r-1 - 2.2n ^r-2 + ... + (n-2)r]

The coefficient of n ^r-k on this left hand side is [1-k] - [k+1-2k] = 0 for

(1 \leq k \leq r - 1)

and the coefficient of n ^r is also 0.

The coefficient of n ⁰ is [-r] - [-2(r)] = r and hence this expression is equal to r as required.