Eli5: Why are there only 88 narcissistic numbers in base 10?
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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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