To elaborate on what Cangurino wrote:

http://en.wikipedia.org/wiki/Modular_arithmetic

we'll use this % as the modulo sign for now...

We have an equation ()

sum(all hat numbers) % n = [0, 1,....,n-2, n-1] (meaning this number can be anywhere between 0 and n-1, and there is n different numbers here)

So if you know the total sum and do % n you get a number, S. But if you have all other hats

sum(other hats) and you know what S is supposed to be you can calculate the number of your hat.

with 5 people S can be one of the numbers 0, 1, 2, 3, 4. Thus we give each person a value S, easiest is giving n1(person 1): S = n-1 (n still being total persons).

So one of the persons will always have the right value of S thus being able to solve what his own hat numbers is correctly.

Lets say S = 3 happens to be right here. (Remember with 5 people S will ALWAYS be between 0 and 4 and we only use whole numbers /integers)

Guy4 looks at other hats and sum it up to 14.

14 % 5 = 4, he needs S to be 3 thus his own hat is 4

18 % 5 = 3.

Not sure if it made it any clearer... beautiful problem and solution imo...

Manhole cover are placed over shafts that lead into underground tunnels. Imagine if you had a square cover, it could be dropped accidentally down the hole. Apparently there is a city that used triangular manhole covers however the problem still exists - someone with butterfingers could drop the cover down the hole. Round covers ensure that due to the diameter of the cover it cannot be dropped down the hole. There are other issues such as protection from sharp edges, autos, etc, but this is the key reason.

Assign one person in the group to be Person A. Person A shall always guess his number correctly. Before the game begins, all participants other than A will agree to observe the number on A's hat. A will stand in the center of the room. Everyone will agree to stand close to A in a number than equals the number on A's hat. Or do something that separates them apart from the group. Basically, they all agree to be "counters" for A. A can then be guaranteed to guess correctly 100% of the time. In the case that A's hat reads 0, everyone turns their back to A.

Let's see if we solve {1,1,2}

0: (0-1-2)%3 = (-3)%3 = 0 -> wrong
1: (1-1-2)%3 = (-2)%3 = 1 -> right
2: (2-1-1)%3 = 0%3 = 0 -> wrong

The following is maybe easier to execute (but equivalent mathematically)

n is the number of players
k is the number assigned to you a priori
1) Take the sum of the numbers you see;
2) Subtract k;
3) Take remainder with respect to n;
4) Subtract from n (unless the result was 0; in that case just guess 0);
5) Take the result as your guess.

So, Player 1 in this case does

1) sum is 3;
2) 3-1 = 2;
3) 2%3 = 2;
4) 3-2 = 1
5) guess is 1