In the arm-wrestling competition, every player plays against every other player twice. Every player plays the same number of games. There are a total of 182 games played. How many players are there?
Assuming if each player does only one game,
When the total number of players is 1, there is 0 game.
When the total number of players is 2, there is 1 game.
When the total number of players is 3, there are 1 + 2 games.
When the total number of players is 4, there are 1 + 2 + 3 games.

Note that the total number of players is 1 more that the last digit in the string of addition.

Let the last digit in the string of addition be 1 u.

Every player plays against every other player twice. 
Total number of games = String of addition x 2.

(1 + 2 + 3 + ... + 1 u) x 2 = 182
(1 + 2 + 3 + ... + 1 u) = 182 ÷ 2
(1 + 2 + 3 + ... + 1 u) = 91

Formula:
Sum of numbers
= 1 + 2 + 3 + ...+ last digit
= ^{last digit x (last digit + 1)}₂

^{last digit x (last digit + 1)}₂ = 1 + 2 + 3 + ...+ last digit
^{1 u x (1 u + 1)}₂ = 91
1 u x (1 u + 1) = 91 x 2
13 x 14 = 182

1 u = 13

Last digit in the string of addition is 13.

Number of players = 13 + 1 = 14

Answer(s): 14