|San José State University|
& the Gateway
to the Rockies
Concerning Weighted Digit Sums
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
The weighted digit sum (WDS) of mn is ha+b where the weight h for m is 10−m.
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.
but in terms of the above process
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.
The process does not work for m≤4 because they have more than one single digit multiple.