﻿ Remainders and the Weighted Digit Sums of Numbers
San José State University

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Thayer Watkins
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Remainders and the
Weighted Digit Sums of Numbers

Suppose the digits of any number are summed and that summation procedure is repeated until the sum is a single digit. That single digit is called the digit sum of the number. It is a notable fact that that digit sum is the remainder upon division of the number under consideration by nine. This result can be generalized.

Let B be the base of a number. Let P be the number pnBn+pn-1Bn-1+…+p0 and D a digit less than B. The number P can be expressed as

#### P = DQ + R

where the remainder R is less than D.

A weighted sum sequence for any number S=snBn+sn-1Bn-1+…+s0 is defined as follows:

#### w0 = snand wi = wi-1h + sn-i

where the weight h is (B−D).

The weighted sum of the digits of any number S is the value of wn, 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.

In a previous study the following theorem is proved.

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.

Consider a number P=DQ+R. Suppose the weighted summation is applied to the number DQ and the result is U. From the theorem U is a multiple of D, say mD. If the weighted summation is applied to P=DQ+R the sequence is the same as for DQ except in the last step and perhaps the next-to-last step. If it differs only in the last step then the weighted digit sum W for P is equal to U+R and hence mD+R, This would mean that the remainder R of division of P by D is equal to the division of W by D.

Consider now the case in which weighted summation results in D+R, but D and R are such that their sum exceeds B, say D+R=B+C. The next weighted sum will give h+C. That means W=h+C. But B+C=D+R implies

#### B−D + C = R but B−D is h hence h + C = R but h+C=W so R = W

This was a special case. More generally the next-to-last step gives mD+R but mD and R are such that their sum exceeds B, say mD+R=kB+C, where k≤m. Thus

#### kB − kD + C = (m−k)D + R or, equivalently kh + C = (m−k)D + R

But the last step in the weighted summation is kh+C so W=(m−k)D + R. Therefore R, the remainder for P divided by D, is the same as the weighted digit sum W of P divided by D. So in all cases this holds true, but in some cases there is the simple result that R=W.

## Illustrations

Let D=8 for B=10. Thus the weight h is 2. Consider the number 45. The weighted sum of its digits is 2·4+5=13. The weighted sum of 13 is 2·1+3=5. Indeed the remainder of 45 upon division by 8 is 5.

Now consider the number 108. The weighted sum of its digits is 2·(2·1+0)+8=4+8=12. The weighted sum of the digits of 12 is 2·1+2=4 and indeed the remainder of 108 upon division by 8 is 4.

Let D=7 for B=10. Thus the weight h is 3. Consider 39. The weighted sum of its digits is 3·3+9=18. The weighted sum of the digits of 18 is 3·1+8=11. The weighted sum for 11 is 3·1+1=4.The remainder for 39 upon division by 7 is 4.

Consider a case for which D<B/2. Let D=3 so h=7. Take P=28. The weighted sum of its digits is 7·2+8=22. The weighted sum for 22 is 7·2+2=16. Then for 16 it is 7·1+6=13. For 13 it is 7·1+3=10 and for 10 it is 7·+0=7. The remainder for division of 28 by 3 is 1. The weighted digit sum of 7 represents 2·3+1. The correct remainder could have been found by finding the remainder upon division of W by D.

Consider another similar case. Let D=4 so h=6. The consider P=34. Its weighted sum is 6·3+4=22. The weighted sum of 22 is 6·2+2=14. The weighted sum of 14 is 6·1+4=10 and that of 10 is 6·1+0=6. Now the remainder upon division of 6 by 4 is 2. And indeed the remainder of 34 upon division by 4 is 2.