Sum of two dice rolls divisible by 3 How many ways are there to roll two distinct dice to yield a sum evenly divisible by 3?
I am having trouble with this one. I know there are 36 possible outcomes, but how would I know which are divisible by 3. Some hints would be appreciated.
 A: Modulo $3$, a die rolls a $0,1,$ or $2$ with equal probability $1/3$.
The ways of summing these to a multiple of three are $0 + 0$, $1 + 2$, and $2 + 1$, i.e. 3 ways.  There are $3 \times 3 = 9$ total possibilities for the two dice, which makes the probability of getting a multiple of three $3/9 = 1/3$.
Since there are $6 \times 6 = 36$ total dice rolls and $1/3$ of those are a multiple of three, the number which are divisible by three is $(1/3)(36) = \boxed{12}$.
A: Ways of getting 3: 12, 21  
Ways of getting 6: 15, 51, 24, 42, 33 
Ways of getting 9: 36, 63, 45, 54 
Ways of getting 12: 66  
That's 12 ways (which happens to be 36/3). Does this mean there are 36/4 = 9 different ways of getting a sum divisible by 4?  
Ways of getting 4: 13, 31, 22 
Ways of getting 8: 44, 35, 53, 26, 62
Ways of getting 12: 66
That's 9 ways. Well well.... 
Divisible by 5:
5: 14, 41, 23, 32  
10: 55, 46, 64, 
7 ways = $ round( \frac{36}{5} = 7.2 )$  
Divisible by 8:
8: 44, 35, 53, 26, 62  
5 ways = $ round( \frac{36}{8} = 4.5 )$  
But divis by 10 is 3 ways = $ \lfloor \frac{36}{10} = 3.6 \rfloor $ so is there no general formula? 
A: Throw the first die. What must be thrown with the second die to get a sum divisible by 3? In all cases there are 2 possibilities. There is a probability of 2/6=1/3 that this occurs. So in 12 of the 36 possibilities the sum is divisible by 3.
