Some interesting aspects of digit sum arithmetic are shown in Quotient 9. Digit sum arithmetic is based upon a property of polynomials so it would apply to numbers with any base not just other integers beside 10, such as a fractional, irrational or transcendental number. The general proof is given Sum of Digits.

Let N_B be a number N represented to base B. That is to say, and N_B=c_nBⁿ+c_n-1B^n-1+…c₁B+c₀, where c_i for i=0 to n belong to a set called digits. For simplicity the set of digits here will be taken to the digits for the base 10. Let S(N_B) be the sum of the coefficients of N_B.

Theorem:

N_B−S(N_B) is an exact multiple of (B−1).

This means that the quotient of [N_B−S(N_B)] divided by (B−1) is a number Q_B where the powers of B run from (n-1) down to 0.

Illustration using B=3/2.

Let N_3/2 = 231_3/2=2(3/2)²+3(3/2)+1=10₁₀. S(N_3/2=2+3+1=6 and .N_3/2−S(N_3/2) = 10−6 = 4. B−1=1/2. So N_3/2−S(N_3/2) divided by B−1 is equal to 8 so N_3/2−S(N_3/2) is an exact multiple of B−1.

Another illustration using B=3/2. Let N_3/2 = 312_3/2=3(3/2)²+1(3/2)+2=41/4=10+1/4. S(312_3/2)=6. Then [312_3/2−S(312_3/2)]=4+1/4=2+(9/4)= 1(3/2)²+0(3/2)+2=102_3/2. Division by (B−1)=1/2 produces 204_3/2. Thus Q_3/2=204 _3/2 and hence [N_3/2−S(N_3/2)] is an exact multiple of. (B−1)

Illustration using B=√2. Let N_√2=221_√2=2*2+2√2+1. Then N_√2−S(N_√2) is equal to 2√2. Consider B−1=√2−1. Then 2√2/(√2−1)=C(√2+1)/(2−1)=4+2√2 = 2√2 + 4. Thus Q=24_√2.

Illustration using B=π. Let N_π=3π²+2π+4. Then S(N_π)=9. But

N_π−S(N_π) = 3(π²−1)+2(π−1)+4−4
= (3(π+1)+2)(π−1)=(3π + 5) = 35_π