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:
- $1, 81, 1458, 1729$
This sequence is A110921 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
Proof
| \(\ds 1\) | \(=\) | \(\ds 1 \times 1\) |
| \(\ds 81\) | \(=\) | \(\ds 9 \times 9\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 9 \times \paren {8 + 1}\) |
| \(\ds 1458\) | \(=\) | \(\ds 81 \times 18\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 81 \times \paren {1 + 4 + 5 + 8}\) |
| \(\ds 1729\) | \(=\) | \(\ds 91 \times 19\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 91 \times \paren {1 + 7 + 2 + 9}\) |
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:
| \(\ds n\) | \(\equiv\) | \(\ds S S'\) | \(\ds \pmod 9\) | Equal Numbers are Congruent | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds S\) | \(\equiv\) | \(\ds S^2\) | \(\ds \pmod 9\) | Congruence of Sum of Digits to Base Less 1 | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds S\) | \(\equiv\) | \(\ds 0 \text { or } 1\) | \(\ds \pmod 9\) |
Suppose $n$ is a $d$-digit integer.
Suppose $d \le 4$. Then $S \le 9 d \le 36$.
The values of $S S'$ are:
| \(\ds 1 \times 1\) | \(=\) | \(\ds 1\) | and we have $1 = 1$ | |||||||||||
| \(\ds 9 \times 9\) | \(=\) | \(\ds 81\) | and we have $8 + 1 = 9$ | |||||||||||
| \(\ds 10 \times 01\) | \(=\) | \(\ds 10\) | and we have $1 + 0 \ne 10$ | |||||||||||
| \(\ds 18 \times 81\) | \(=\) | \(\ds 1458\) | and we have $1 + 4 + 5 + 8 = 18$ | |||||||||||
| \(\ds 19 \times 91\) | \(=\) | \(\ds 1729\) | and we have $1 + 7 + 2 + 9 = 19$ | |||||||||||
| \(\ds 27 \times 72\) | \(=\) | \(\ds 1944\) | and we have $1 + 9 + 4 + 4 \ne 27$ | |||||||||||
| \(\ds 28 \times 82\) | \(=\) | \(\ds 2296\) | and we have $2 + 2 + 9 + 6 \ne 28$ | |||||||||||
| \(\ds 36 \times 63\) | \(=\) | \(\ds 2268\) | and we have $2 + 2 + 6 + 8 \ne 36$ |
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.
Aiming for a contradiction, suppose $d \ge 5$.
Suppose $d$ is a $k$-digit integer.
We have $k = 1 + \floor {\log d}$.
We show that $2 k + 2 \le d - 1$:
- $d - 1 \ge 4 = 2 k + 2$
- 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:
| \(\ds d - 2 \log d - 5\) | \(>\) | \(\ds \map f 7\) | $f$ is strictly increasing for all $d \ge 7$. | |||||||||||
| \(\ds \) | \(>\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds d - 1\) | \(>\) | \(\ds 2 \log d + 4\) | |||||||||||
| \(\ds \) | \(\ge\) | \(\ds 2 \floor {\log d} + 4\) | Definition of Floor Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 k + 2\) | $k = 1 + \floor {\log d}$ |
- 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.
$\blacksquare$
Historical Note
David Wells reports in his Curious and Interesting Numbers, 2nd ed. of $1997$ that this result was discussed by Charles Wilderman Trigg in volume $9$ of Journal of Recreational Mathematics, page $44$, but it has not been possible to corroborate this.
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $1729$