﻿ The Net Sum of a Decimal Representation of a Number Resulting From an Alternation of Addition and Subtraction
San José State University

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Thayer Watkins
Silicon Valley
USA

The Net Sum of a Decimal Representation
of a Number Resulting From an
and Subtraction

## Background

The digit net sum of a number involves adding and subtracting successive digits of a number, Take the number 137. Its digit net sum is evaluated from the right as 7−3+1=5. The sum of its digits Is 1+3+7=11. Its digit sum is the sum of the digits of that number; i.e., 2=1+1. The digit sum of a number has the property of being equal the remainder of the number upon division by 9; i.e., 137 = 15*9 + 2. Provision has to be made for the digit sum being equal to 9 whereas the remainder is 0.

The digit net sum of a number also has an interesting property. It may be the remainder for the number upon division by 11. The digit net sum of 137 was found to be 5 and 137=12*11+5. Things are a bit more complicated than for the digit sum and 9. Consider 2001. Its digit net sum is 1−0+0−2 which is −1 and 2001=187*11+10, but 2001=188*11−1. The difference between 137 and 2001 is that highest power of 10 in 137 is the second, whereas in 2001 it is the third power. Let N be a decimal number and let n be the highest power of ten it contains. Let ρ(N) be its digit net sum. What can be said is that

#### N−ρ(N) is exactly divisible by 11

Another way of expressing the relationship is

#### N = ρ(N) (modulo 11)

In the case of the number 2001, 10=−1 (mod 11).

## Digit Net Sum and Modulo 11 Arithmetic

Consider the numbers 42 and 74. Their digit net sums are −2 and −3, respectively. The product of 42 and 74 is 3108 and its digit net sum is +6. The product of the digit net sums of 42 and 74 is +6. There is obviously a relationship between the digit net sum of a product and and the product of the digit net sums of numbers being multiplied, but it may be a bit complicated . The remainder of 42 modulo 11 is 9 and 9=−2 (mod 11). The remainder of 74 modulo 11 is 8 and 8=−3 (mod 11). The product of these remainders is 72 which is equal to 6 (mod 11). But the digit net sum of 72 is −5, which is equal to 6 (mod 11). Again, as was said above, there is obviously a relationship between the digit net sum of a product and and the product of the digit net sums of numbers being multiplied, but it could be a bit complicated.

For digit sums the relationship is straight forward. The digit sums of 42 and 74 are 6 and 2=1+1 for 7+4=11. The product of the digit sums is 12 and its digit sum is 3. The sum of the digits of the product 3108 is 12 and its digit sum is 3.

## The Analysis

Whatever relationships hold for digit net sum and modulo 11 arithmetic hold for polynomials over a field, but the analysis will be carried out for polynomials over the nonnegative integers.

Consider the following two multiplications.

#### (k+1)(k³ −k² + k −1) which can be visualized as k4 − k³ + k² − k              k³ −k² + k −1 ______________________________________ k4 −1

Clearly, in general,

#### (k+1)(kn-1−kn-2+ … + (−1)n-2k + (−1)n-11 = kn + (−1)n-11 = kn − (−1)n1

Thus for n even (k+1) is an exact factor of kn −1 and for n odd it is an exact factor of kn + 1.

Consider a polynomial

#### P(C, k) = cnkn + cn-1kn-1 + … + c1k + c01

where C represents the sequence of coefficients {cn, … c0}.

Now define

Then

#### (P(C, k)) − ρ(P(C, k)) = cn[kn−(−1)n] … c1(k+1)

(k+1) is a factor of each term on RHS and thus is a factor of the sum. This means that

#### (P(C, k)) − ρ(P(C, k)) = (k+1)P(D, k) and hence (P(C, k)) = (k+1)P(D, k) + ρ(P(C, k))

Thus ρ(P(C, k)) is a remainder for division of the polynomial by (k+1) but if ρ(P(C, k)) is negative then the usual notion of the remainder is [(k+1) + ρ(P(C, k))]. Thus

## The Product of Polynomials

Let P(C, k) and P(D, k) be two polynomials in k. Then

#### ρ(P(C, k)*P(D, k)) = P(C, −1)*P(D, −1) = ρ(P(C, k)*ρ(P(D, k))

But if the RHS is not a single digit the digit net sum of it must be computed.

## The Sum of Polynomials

Again let P(C, k) and P(D, k) be two polynomials in k. Then

#### ρ(P(C+D, k)) = ρ(P(C, k)+P(D, k)) = ρ(P(C, −1)+P(D, −1)) = ρ(ρ(P(C, k)+ρ(P(D, k)))

For the problem under consideration k=10 and (k+1) is 11.

## Modulo 11 Arithmetic

The Product of Integers Modulo 11
0 1 2 3 4 5 6 7 8 9 10
0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6 7 8 9 10
2 0 2 4 6 8 10 1 3 5 7 9
3 0 3 6 9 1 4 7 10 2 5 8
4 0 4 8 1 5 9 2 6 10 3 7
5 0 5 10 4 9 3 8 2 7 1 6
6 0 6 1 7 2 8 3 9 4 10 5
7 0 7 3 10 6 2 9 5 1 8 4
8 0 8 5 3 10 7 4 1 9 6 3
9 0 9 7 5 3 1 10 8 6 4 2
10 0 10 9 8 7 6 5 4 3 2 1

If xy=1 (mod 11) then x and y are multiplicative inverses of each other. In each column and each row there is an entry of 1. Thus each integer has a multiplicative inverse. The multiplicative inverse of 2 modulo 11 is 6. Thus multiplication by 2 modulo 11 is reversed by subsequent multiplication by 6 modulo 11. For example, 137 is equal to 5 modulo 11 and 2*137=274=10 (mod 11). But 6*274 equals 1644 which is equivalent to 5 modulo 11.

Arithmetic modulo 11 is completely consistent. It constitutes a mathematical field because 11 is a prime number.

(To be continued.)