How to determine whether a number is divisible by 3

Examples :
45
4 + 5 = 9
9 / 3 = 3

2550
2 + 5 + 5 + 0 = 12
12 / 3 = 4

1290363
1 + 2 + 9 + 0 + 3 + 6 + 3 = 24
24 / 3 = 8
or, if you don't like dividing 24 by 3,
24
2 + 4 = 6
6 / 3 = 2

Voila!

Proof of the above method:

Consider each digit seperately. The digit in the '1's place (d₀) is just exactly what it is. The digit in the '10's place (d₁) has the value (10 * d₁) = ((3 * 3 * d₁) + d₁) = ((3 * N₁) + d₁) for N₁ equal to some integer (specifically, (3 * d₁)). Now, A is divisible by 3 if and only if (A - (3 * N₁)) is divisible by 3. So the number ((10 * d₁) + d₀) is divisible by 3 iff (((10 * d₁) + d₀) - (3 * N₁)) is divisible by 3, i.e. iff (d₀ + d₁) is divisible by 3. The same operation can be performed for all the remaining digits, (100 * d₂) = ((3 * 33 * d₂) + d₂), N₂ = (33 * d₂), and in general ((10 ^ i) * d_i) = ((3 * 3 * f(i) * d_i) + d_i), N_i = (3 * f(i) * d_i) where f(i) is the value represented by i '1' digits in a row in decimal notation.

This shows that not only can you find whether a number is divisible by 3, but exactly what number it is modulo 3 - if you end up with 0, 3, 6, or 9 then the original number is 0 modulo 3, if you end up with 1, 4, or 7 then the original number is 1 modulo 3, and if you end up with 2, 5, or 8 the original number is 2 modulo 3.

The same argument can be used to check if a number is divisible by 9, or to check what value a number is modulo 9 (use (9 * d₁), N₁ = d₁, and (9 * 11 * d₂), N₂ = (11 * d₂), (9 * f(i) * d_i), N_i = (f(i) * d_i)).