I’m not sure that there’s an ELI5 answer, but here’s a go.

A number (let’s call it the ‘input’) is narcissistic when it equals the sum of each of its digits, each raised to the power of its length (let’s call that the ‘output’).

On a graph, the line of available numbers, the input, is straight – x=1, y=1 and so on.

The second line, plotting the output is wiggly – all graphs of polynomial outputs are. It is also slanted upwards. Every time the output line crosses the input line, we have a narcissistic number.

However, there comes a point where the output line is above the input line, and *never comes back below it* – ie, above a certain point, the output is always bigger than the input. From that point on, there are no more narcissistic numbers.

Edit: sum not product. As u/chromotron points out, this would be very easy if it was the product!

If you count 0, there are 89, 88 if you don’t count 0.

This isn’t quite ELI5, but nor is the concept of narcissistic numbers, so I won’t tell if you don’t..

Suppose you have an n-digit number, and it’s narcissistic. Since it’s an n-digit number, it’s at least 10^(n-1). However, the sum of its digits is at most nx9^(n). For n bigger than 60, it turns out that n x 9^(n) is always less than 10^(n-1). So there can’t be any narcissistic numbers with 60 or more digits.

For fewer digits, well, very loosely, if you pick a random number, the “chance” of the sum coming out exactly right is tiny – about 1 in 10^(n). But you get about 10^(n) tries, since there’s roughly that many n digit numbers, so you’ll get about 1 narcissistic number with n digits. These are *extremely* loose and hand-wavy calculations, but all they show is that 88 is roughly reasonable.

Lets write N(x) for the “narcissism” of x, defined as as the sum of the k-th powers of the decimal digits of x, where k is the decimal length of x. Than by definition a number is narcissistic iff N(x) = x.

Lets say we have such a narcissistic x with k digits. note that 10^(k-1) is the smallest number with k digits, a single one followed by k-1 many zeros. Hence we know that 10^(k-1) ≤ x.

On the other hand, N(x) is at most 9^k · k, one 9^k per digit. Combining everything we observe that we must have

10^(k-1) ≤ x = N(x) ≤ 9^k · k.

But the total inequality 10^(k-1) ≤ 9^k · k is wrong for sufficiently large k. To argue this, rewrite it as (10/9)^k ≤ 10·k, now the left hand side grows exponentially, the other side does not, hence it will at some point be overtaken.

Slightly more careful checks show that this happens around k=60, therefore a narcissistic number cannot have more than 60 digits. All left to do is checking the ~10^60 remaining numbers “by hand”. To do this sanely and within a lifetime, a few more arguments can be thrown in to narrow the search down, ultimately leading to those 88 numbers.

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