Theorem: Numbers that Factorise into Sum of Digits and Reversal

The following positive integers can each be expressed as the product of the sum of its digits and the reversal of the sum of its digits:
:$1, 81, 1458, 1729$
{{OEIS|A110921}}

Proof:
{{begin-eqn}}
{{eqn | l = 1
      | r = 1 \times 1
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 81
      | r = 9 \times 9
}}
{{eqn | r = 9 \times \paren {8 + 1}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 1458
      | r = 81 \times 18
}}
{{eqn | r = 81 \times \paren {1 + 4 + 5 + 8}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 1729
      | r = 91 \times 19
}}
{{eqn | r = 91 \times \paren {1 + 7 + 2 + 9}
}}
{{end-eqn}}
Let $n$ be a positive integer.
Let $S$ the sum of its digits and $S'$ be the reversal of the sum of its digits.
We wish to determine integers $n$ that satisfy:
:$n = S S'$
From this we have:
{{begin-eqn}}
{{eqn | l = n
      | o = \equiv
      | r = S S'
      | rr = \pmod 9
      | c = Equal Numbers are Congruent
}}
{{eqn | ll = \leadsto
      | l = S
      | o = \equiv
      | r = S^2
      | rr = \pmod 9
      | c = Congruence of Sum of Digits to Base Less 1
}}
{{eqn | ll = \leadsto
      | l = S
      | o = \equiv
      | r = 0 \text { or } 1
      | rr = \pmod 9
      | c = 
}}
{{end-eqn}}
Suppose $n$ is a $d$-digit integer. 
Suppose $d \le 4$. Then $S \le 9 d \le 36$.
The values of $S S'$ are:
{{begin-eqn}}
{{eqn | l = 1 \times 1
      | r = 1
      | c = and we have $1 = 1$
}}
{{eqn | l = 9 \times 9
      | r = 81
      | c = and we have $8 + 1 = 9$
}}
{{eqn | l = 10 \times 01
      | r = 10
      | c = and we have $1 + 0 \ne 10$
}}
{{eqn | l = 18 \times 81
      | r = 1458
      | c = and we have $1 + 4 + 5 + 8 = 18$
}}
{{eqn | l = 19 \times 91
      | r = 1729
      | c = and we have $1 + 7 + 2 + 9 = 19$
}}
{{eqn | l = 27 \times 72
      | r = 1944
      | c = and we have $1 + 9 + 4 + 4 \ne 27$
}}
{{eqn | l = 28 \times 82
      | r = 2296
      | c = and we have $2 + 2 + 9 + 6 \ne 28$
}}
{{eqn | l = 36 \times 63
      | r = 2268
      | c = and we have $2 + 2 + 6 + 8 \ne 36$
}}
{{end-eqn}}
Among these values, only $1, 81, 1458, 1729$ have the desired property.
We claim that there are no integers with more than $4$ digits with this property.
{{AimForCont}} $d \ge 5$.
Suppose $d$ is a $k$-digit integer.
We have $k = 1 + \floor {\log d}$.
We show that $2 k + 2 \le d - 1$:
:$d = 5,6$ are single-digit integers, so $k = 1$.
:$d - 1 \ge 4 = 2 k + 2$
:For $d \ge 7$, consider the function $\map f d = d - 2 \log d - 5$.
:Then $\map f 7 > 7 - 2 - 5 = 0$.
:We also have $\map {f'} d = 1 - \dfrac 2 {d \ln 10} > 1 - \dfrac 1 d$,
:so $\map {f'} d > 0$ for all $d \ge 7$.
:By Real Function with Strictly Positive Derivative is Strictly Increasing, $f$ is strictly increasing for all $d \ge 7$.
:Then:
{{begin-eqn}}
{{eqn | l = d - 2 \log d - 5 
      | o = > 
      | r = \map f 7
      | c = $f$ is strictly increasing for all $d \ge 7$.
}}
{{eqn | o = > 
      | r = 0
      | c = 
}}
{{eqn | ll = \leadsto
      | l = d - 1
      | o = >
      | r = 2 \log d + 4
}}
{{eqn | o = \ge
      | r = 2 \floor {\log d} + 4
      | c = {{Defof|Floor Function}}
}}
{{eqn | r = 2 k + 2
      | c = $k = 1 + \floor {\log d}$
}}
{{end-eqn}}
:So we have $2 k + 2 \le d - 1$ for all $d \ge 5$.
$9 d$ has not more than $k + 1$ digits.
Since $S \le 9 d$, $S$ cannot have more digits than $9 d$.
We also have that $S'$ cannot have more digits than $S$.
Therefore we have $S, S' < 10^{k + 1}$. 
Then $n = S S' < 10^{2 k + 2} \le 10^{d - 1} \le n$, which is a contradiction.
The result follows by Proof by Contradiction.
{{qed}}