Narcissistic Numbers

There are more: for example, any single digit number, or $370$ , or (of course) $371$ . We'll show there are only finitely many more.

The largest possible total of $n$ digits raised to the $n^{\text{th}}$ power is when all the digits are $9$ , giving a total $n \cdot 9^n$ .

This total itself has $d(n)=\left \lceil \log_{10} \left( n \cdot 9^n \right) \right \rceil$ digits. Since we're not concerned with the exact number of solutions, it's enough to show that for some $N$ we have $d(n)<n$ for all $n \ge N$ .

Now, $\begin{aligned} d(n) &< \log_{10} \left( n \cdot 9^n \right)+1 \\ &= 1+\log_{10} n + n\log_{10} 9 \end{aligned}$

It's easy to check that we can take $N=61$ to satisfy the above condition. Hence there is a finite limit to the length of narcissistic numbers, and so there are only finitely many of them.

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We don't need to be too precise - we only want to show there are not infinitely many narcissistic numbers. We can use some rough bounds; first, note that the function $f(x)=\frac{\log_{10} x}{x}$ is decreasing for $x>e$ (easy enough to check by differentiating).

So for all $n>85$ , $f(n)<f(85)<\frac{1}{44}$ . Also, $\log_{10} 9 < \frac{21}{22}$ . Hence for $n>85$ , $\begin{aligned} d(n)&<1+\frac{n}{44}+n\frac{21}{22} \\ &=1+\frac{43n}{44} \\ &=n \left(\frac{1}{n}+\frac{43}{44}\right) \\ &<n \left(\frac{1}{44}+\frac{43}{44}\right) \\ &=n \end{aligned}$

so without any tricky analysis we've proved that $N\le 86$ , again showing there are finitely many narcissistic numbers.

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Final note: there are, in fact, $89$ narcissistic numbers base $10$ (or $88$ if you exclude zero), and you can find out more about them here