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The Weighted Digit Sums
of Multiples of Digits

It is a notable fact that the sum of the digits of any positive multiple of nine is a multiple of nine, If the summation procedure is repeated
until the sum is a single digit that digit is nine. That single digit is called the digit sum of the number. Thus the digit sum of any
multiple of nine (other than zero) is nine. This material is a generalization of that notable property.

Let B be the base of a number. Let the number M=m_{n}B^{n}+m_{n-1}B^{n-1}+…+m_{0} be a multiple of a
digit D. Thus M=DQ, where Q is an integer.

A weighted sum sequence for M is defined as follows:

w_{0} = m_{n} and
w_{i} = w_{i-1}h + m_{n-i}

The weight h is (B−D).

The weighted sum of the digits of M is the value of w_{n}, the final term of the sequence.
When this process is repeated until the result is a single digit W that single
digit is known as the weighted digit sum.

Theorem:

The weighted digit sum W of a multiple of D is a multiple of D. Therefore for D≥B/2,
W is equal to D.

Illustrations

If D=9 and B=10 then the weight h=1 and the sum of the digits of multiples of 9 are all multiples of 9 and if the processes of summing
digits is carried forward the end result is always 9.

Let D=8 for B=10. Thus the weight h is 2. Consider the multiples of 8. For 16, W=2·1+6=8. For 24, W=2·2+4=8.
Jumping to 56=7·8. The weighted sum of the digits of 56 is 2·5+6=16. The weighted sum of the digits of 16 is,
as already noted, 8. Now consider 13·8=104. The weighted sum of the digits of 104 is 2(2·1+0)+4 is 8.

Consider now D=7, for which h=3. The weighted sum of the digits of 21 is 3·2+1=7. For 49 the weighted sum of the digits for
is 3·4+9=21, so repeating the process gives W=7.

Now consider D=4, a digit which is less than 10/2. The weight h for D=4 is 6. For 24 the weighted digit sum is 6·2+4=16.
For 16 it is 6·1+6=12 and for 12 it is 6·1+2=8, a multiple of 4.

Proof

The sequence {w_{i} for i=0 to n} can be expressed as

The leading term for each h^{j}=(B−D)^{j} is B^{j}. All of the rest of the terms have D as a factor.
When all of these leading terms are multiplied by their coefficients and combined with the term m_{0} the result is

m_{n}B^{n} + m_{n-1}B^{n-1} + … + m_{0}

This is none other than the number M. All of the other terms have D as a factor. Thus

w_{n} = M + DS

where S is a negative integer. But M is a multiple of D; i.e., M=DQ. Therefore

w_{n} = DQ + DS = D(Q + S)

Thus the weighted sum of the digits of a multiple of a digit D is a multiple of D.