Digit sum arithmetic has surprising properties, so much so that stage performers such as mathemagician Arthur Benjamin uses it as the basis for apparently amazing feats of computation. Benjamin does not mention it. The performer Shakuntala Devi calls it hidden arithmetic. The digit sum of a number is sometimes called its digital root. Digital root is a name so opaque that it is stupid. It gives no hint of what it really is and implies something it it not; i.e., a root being the solution of an equation. In contrast digit sum reveals that it is a sum of the digits of a number which is a single digit.

But digitsum arithmetic is just the application of more general properties of polynomials applied to powers ten or any other base numbering system.

Consider polynomials of the form

N = c_nkⁿ + c_n-1k^n-1 + … + c_mk^m

where n and m are integers, positive or negative, with m<n and k and the coefficients c_j are real numbers. The quantity k will be referred to as the base of the polynomial. The largest power n is called the degree and the smallest power m will be called the minimum power, The difference (n−m) will be called the degree span of the polynomial. An irrational number such as √2 expressed in powers of 10 is a polynomial of infinite degree span. Later the extension of the analysis to complex numbers and beyond will be considered.

Such a polynomial as above may be represented as

N = P(C, k)
where C is
the sequence
{c_n, … c_m}

Let n now be a positive integer.

Lemma: (kⁿ−1) is equal to (k−1)(kⁿ⁻¹ + kⁿ⁻² + … +1)

Proof:

(k−1)(kⁿ⁻¹ + kⁿ⁻² + … +1) =  kⁿ + kⁿ⁻¹ + kⁿ⁻¹ + … + k
                                                              − (kⁿ⁻¹ + kⁿ⁻² + … + k + 1)
  _____________________________________________________
  = kⁿ − 1

Thus, in particular (k-1) is an exact factor of (kⁿ − 1) for all positive n.

Let σ(P(C, k)) denote the sum the coefficients of a polynomial. Then

Theorem 1: (k-1) is an exact factor of [P(C,k) − σ(P(C, k))].

Proof:

[P(C,k) − σ(P(C, k))] = Σ c_jk^j − Σ c_j
= Σ c_j(k^j−1)
(k-1) exactly divides
each (k^j−1)
therefore it exactly divides the sum

Illustration

Let N equal 2k² + 3k +1 so σ(N)=6. Let k equal √2. Then

(k-1) = 0.414213562373095…
N = 9.242640687119285…
N−σ(N) = 3.242640687119285…
(N−σ(N))/(k-1) = 7.828427124746191…
= 2·2 + 2√2 + 1

Sums

Theorem 2: If N=P(C, k) and M=P(D, k) then

N+M=P((C+D), k)
and hence
σ(N+M)=σ(N)+σ(M)

The validity of this theorem is obvious.

Products

Consider the product

M = (ak+b)(ck+d) = ack² + (ad+bc)k + bd = σ(ak+b)·σ(ck+d)

The coefficient sum σ( ) of a polynomial can be obtained by setting its base k equal to 1; i.e., σ(P(C, k) = P(C, 1). Therefore

Theorem 3: If N=P(C, k) and M=P(D, k) then

σ(N·M) = σ(P(C, k)·P(D, k))
= P(C, 1)·P(D, 1)
= σ(N)·σ(M)

Illustration

Let N equal 2k² + 3k +1 and M equal k+1 for k=√2. Then

N = 9.242640687119285…
M = 2.414213562373095…
N·M = 22.31370849898476…
= 2(2^3/2) +5·2 + 4·2^1/2 + 1
hence
σ(N·M) = 12
which is the same as
σ(N)·σ(M) = 6·2

Now consider the remainder of a polynomial upon division by (k-1). A polynomial N of finite degree span can be expressed as

N = (k-1)M + R

where M is a polynomial of degree one less than that of N. The deree of R is equal to the minimum power of N.

Taking the coefficient sums of both sides of this equation gives

σ(N) = σ((k-1))σ(M) + σ(R)

But σ((k-1))=0 and hence

Theorem 4: σ(N) = σ(R)

If m is the minimum power of N then R=d₀k^m and hence σ(R)=d₀. The quantity d₀ is what would be called the remainder for division of N by (k-1).

Illustration

Let N equal 2k² + 3. Then M=2k+2 and R=5. Clearly σ(N)=5 and σ(R)=5.

Conclusion

The coefficient sums of the sum and the product of two polynomials of finite degree spans and of the same base are the sum and product, respectively, of the coefficient sums of the polynomials. The difference between a polynomial and its coefficient sum is a multiple of its base less 1. The coefficient sum of a polynomial of finite degree span is equal to its remainder upon division by its base less 1.