The Two Digit Cases

Let m be a digit greater than or equal to 5 but less than or equal to 9. Let mn be a multiple of m such that

mn = 10a + b

The weighted digit sum (WDS) of mn is ha+b where the weight h for m is 10−m.

Thus

WDS(mn) = (10−m)a + b = 10a + b −ma = mn −ma = m(n−a)

For m≤9, a<n so (n−a)≥0 and (n−a)≤n. Thus application of WDS reduces the multiple of m until n=1 and hence WDS=m.

Illustation with m=8 and then h=2. Consider n=4 and hence 8*4=32.

WDS(32) = 2*3 + 2 = 6 +2 = 8

but in terms of the above process

WDS(32) = (10−8)3 + 2 = 10*3 + 2 −8*3 = 32 −8*3 = 32 − 24 = (4 −3)*8 = 1*8 = 8

For a multiple of m having more than two digit the WDS process is applied to the left-most two digits until the first left-most digit is eliminated. The process continues with the new two left-most digits. At each stage the result is a multiple of m until the last stage in which the multiple of m is 1*m.

For example. for m=8 and n=14, mn=112. For m=8 h=2.

WDS(11) = 2*1 + 1 = 3
WDS(32) = 2*3 + 2 = 8

The process does not work for m≤4 because they have more than one single digit multiple.