Fooling around with factorials -- part 2 -- with thanks to Alisa Meier

Number Theory Level 4

Let $S(n)$ denote the number of digits of $n$ . Then find the sum of all $n$ 's that satisfy the following equation:

$\large S(n!) + 1 = n$

1 solution

Andrey Bessonov
Aug 23, 2015

S(n!)+1 = log10(n!)

So, it`s grow slower then n, for n<10, and faster for n>10. We need to find two sequences of right numbers (for n<10 and n>10)..

For n<10 answer is {2}

For n>10 answer is {19...21}

Final answer: 2 + 19 + 20 + 21 = 62