Elegant number limited

A number is called "elegant" if it is equal to the sum of its digits raised to the consecutive powers in decimal respresentation. In other words, if $S = a_1 a_2 \dots a_n$ is elegant then $S = a_1^1 + a_2^2 + \dots + a_n^n$

For example, $89 = 8^1 + 9^2$ is an elegant number. Also, $$2646798 = 2^1 + 6^2 + 4^3 + 6^4 + 7^5 + 9^6 + 8^7$$ is an elegant number

Prove or disprove: there exists an infinite number of elegant numbers

Solution

We shall prove that there exists only a finite number of elegant numbers because for large enough $N$, it is impossible for an $N$-digit elegant number to exist.

If $S = a_1\dots a_N$ is elegant then $S \geq 10^{N-1}$ because $S$ is an $N$ digit number. But at the same time: $$S = a_1^1 + \dots + a_N^N \leq 9^1 + \dots + 9^N = \frac{9^{N+1}-1}{8}$$ So we have: $$10^{N-1} \leq \frac{9^{N+1}-1}{8}$$ The function on the LHS grows faster than the RHS because the exponentiation base is larger. So this inequality ceases to be true for large enough $N$. In fact: $$\iff (10/9)^{N+1} \leq 12.5$$ $$\iff N \leq \frac{\log (12.5)}{\log (10/9)} -1 \sim 22.9$$ So for $N = 23$ we already find that there cannot exist a 23-digit elegant number