Consider a non-negative integer expressed in Base 10 notation. Its DigitSum is the sum of its digits computed iteratively until a single digit result is obtained. For example, the sum of the digits of 386 is 3+8+6=17. Then the sum of thedigits of 17 is 8. Thus 8 is the DigitSum of 386. The DigitSum function of an integer Z will be denoted as D₁₀(Z) Thus D₁₀(386)=D₁₀(17)=8.

One remarkable fact about the DigitSum function is that it is equal to the remainder for a number upon division by 9, with the provision that 9 and 0 are equivalent. Let the remainder upon division by 9 for a non-negative integer Z be denoted as r₉(Z). The proposition is that

D₁₀₁₀(Z) = r₉(Z)
if D₁₀(Z) belongs to {1, 2, … 8}
and
if D₁₀(Z)=9 then r₉(Z)=0.

There is nothing special about the base 10 notation. For any base B the DigitSum of any integer is equal to its remainder upon division by (B-1), with the provision that a DigitSum of (B-1) is equivalent to a remainder of 0 upon division by ((B-1).

The proof will be carried out for numbers in base 10 notation, but the extension to an arbitrary base B is obvious.

First some of the properties of remainder arithmetic must be considered. The remainder for a number N for division by M is an integer r between 0 and (M-1) such that N=Mk+r for some integer k.

Lemma: The remainder of N for division by M is the same as the remainder of N-hM for any hM≤N.

Proof:

If N-hM = kM + r for 0≤r<M then also

N = (h+k)M + r

Let N₁ and N₂ with remainders of r₁ and r₂, respectively. Then

r₉(N₁+N₂) = r₉(r₁+r₂)

Likewise

r₉(N₁*N₂) = r₉(r₁r₂)

Now consider a two digit number n=ab=10a+b, where a and b belong to the set {0, l, 2, …, 9}.

Then

r₉(n) = r₉(10a+b)= r₉(r₉(10a)+r₉(b))
but r₉(b) = b and
r₉(10a) = r₉(10)*r₉(a) = 1*r₉(a) = a
and thus
r₉(n) = r₉(a+b)

But (a+b) is equal to some 10c+d and thus

r₉(n) = r₉(c+d)

This process must terminate with some digit q. This digit q is D₁₀(n).

If D₁₀(n) belongs to the set {0, 1, 2, …, 8} then
r₉(n)=D₁₀(n).
If D₁₀(n)=9 then
r₉(n)=r₉(9)=0.

The General Case

Let n be an m digit number. Then n=Σ10ⁱa_i for i=0 to m and hence'

r₉(n) = r₉(Σa_i)

But Σa_i is equal to some number Σ10^jb_j and so

r₉(n) = r₉(Σb_j)
and so on down to a single digit D₁₀(n)

Thus

r₉(n) = r₉(D₁₀(n))
and hence
if D₁₀ belongs to {0, 1, 2, …, 8} then
r₉ = D₁₀
and if D₁₀=9 then r₉ =0

A More General Case

Let B be the base of a notation for integers. Thenfor an integer n

r_B-1(n) = r_B-1(D_B(n))

Thus

if D_B(n) belongs to {0, 1, 2, … (B-2)} then r_B-1(n) = D_B(n)
and r_B-1(n) = 0 if D_B(n)=B-1

The number 36 in base 10 notation is 44 in base 8 notation. This is denoted as 44₈ The DigitSum of 44 is 8=10₈ and hence D₈(44₈)=1. The remainder of 36₁₀=44₈ upon division by 7 is 1, i.e.; r₇(44₈) = 1₈.

Extension to Negative Integers

The DigitSum and remainder functions for a negative integer N can be defined as the negative value of the corresponding functions for the absolute value of N; i.e.,

D_K(N) = sgn(N)D_K(|N|)
r_K(N) = sgn(N)r_K(|N|)

With these definitions all of the previous relationships also hold for negative integers.