To set the sticky bit on a directory, why do the commands
chmod 1777 and
chmod 3777 both work?
Each number (also referred to as an octal because it is base8) in that grouping represents 3 bits. If you turn it into binary it makes it a lot easier.
1 = 0 0 1
3 = 0 1 1
5 = 1 0 1
7 = 1 1 1
So if you did 1777, 3777, 5777, or 7777 you would set the sticky bit because the third column would be a 1. However, with 3777, 5777, and 7777 you are additionally setting other bits (SUID for the first column, and SGID for the second column).
Conversely, any other number in that spot (up to the maximum of 7) would not set the sticky bit because the last column wouldn't be a 1 or "on."
2 = 0 1 0
4 = 1 0 0
6 = 1 1 0
The permissions passed as an argument to chmod are specified as an octal value. Each numeral in the value represents three bits. If three numerals are given, you're setting the read, write and execute bits for the file's owner, group and others (everyone else). If four numerals are given, the leftmost number sets the setuid, setgid and sticky bits. Octal 1 sets the sticky bit. Octal 2 sets the setgid bit. Octal 2 + octal 1 is octal 3 which sets both the setgid bit and the sticky bit.