Let the digital representation of a number n in base K be denoted as n_K. Suppose n_K is a two digit number aK+b. Its weighted digit sum with respect to a digit m is

WDS_m(aK+b) = ha + b
where h=K−m

Thus

WDS_m(n_K) = WDS_m(aK+b) = (K−m)a + b = aK + b −am = n −am

Now suppose n=sm+r, where r is the remainder for the division of n by m and s is a positive integer. Then

WDS_m(n_K) = sm+r −am = (s−a)m + r

That is, WDS_m(n_K) is a smaller multiple of m than n and has the same remainder upon division by m. If the process is continued until the result is less than K the remainder of the division of that result is the same as for the division n by m.

Conclusions

The remainder for the division of a number n by a digit m is the same as the division of the weighted digit sum of that number represented to base K larger than m.