Prove that if n is a positive integer such that the sum of the digits of n in decimal representation is divisible by 3 then n is divisible by 3
This one is sort of the archetype for the rest of them.
Note that
$\displaystyle 10^n \equiv 1 \text{ mod 3}$
for all positive n.
So putting a number x into its decimal representation in base 10:
$\displaystyle x = \sum_n x_n \cdot 10^n$
We see that
$\displaystyle x \equiv \sum_n x_n \cdot 1 \text{ mod 3}$
So if $\displaystyle \sum_n x_n = 0 \text{ mod 3}$ (ie. the sum of the digits of x is divisible by 3), then the number x is divisible by 3.
-Dan