Numbers that Factorise into Sum of Digits and Reversal

Theorem
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:
 * $1, 81, 1458, 1729$

Proof
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:

Suppose $n$ is a $d$-digit integer.

Suppose $d \le 4$. Then $S \le 9 d \le 36$.

The values of $S S'$ are:

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.

$d \ge 5$.

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:


 * 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.