Numeral system information density
Let you want to represent integer values from 0 to 999. You need 3 digits, each has 10 states, 30 states overall. When using binary digits, you need 10 digits with 2 states, 20 states overall, So binary representation is more "dense". In general case, p*ln(N)/ln(p) states is required. This formula has minimum at p=e. However, this is pure theoretical speculation. Numeral system with the base 3 is slightly more dense than that of 2, and 3-based computers were really built, but appeared more complex than binary computers.