{{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}}