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This works for whole numbers. Simply add up the digits in the number. The resulting number should be small enough to divide by three in your head. If that number is evenly divisible by three, so is the original number. I heard this long ago in fourth grade.

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

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) * di) = ((3 * 3 * f(i) * di) + di), Ni = (3 * f(i) * di) 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) * di), Ni = (f(i) * di)).

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