Dice sums

Let us roll the first 3 dice...note that whatever the outcome of the first 2 dice may be there is only one possible outcome from the 4th dice to make the whole thing divisible by 3.So the number of possibilities = $6*6*6*2 = 432$

I took a very simple route. The total number of combinations for $d_1, d_2, d_3, d_4$ are $6^{4}$ . This is where I made an educated guess. $\frac{1}{3}$ of all numbers is divisible by $3$ . This can be extended to the sum of all the $d$ 's. Thus, the number of sums divisible by $3$ is the total combinations divided by $3$ .

So, $6^{4}$ can be rewritten as $2^{4}\times3^{4}$

Thus, $\frac{2^{4}\times3^{4}}{3}$ $=$ $2^{4}\times3^{3}$ $=$ $16\times27$ $=$ $\boxed{432}$

Ok so uhm I don't really use formulas much so I based my answer more on logic. The numbers on the first three dice don't really matter, since the last die is the only one which determines whether the sum becomes a multiple of 3 or not. So in the first three dice, there are 216 ways of how the numbers of the dice end up. Then here, let's consider 3 cases. If the sum of the numbers of the first three dice: -is a multiple of 3, then you only need to add 3 or 6 -is 1 more than a multiple of 3 (like four, for example), you only need a 2 or a 5 -is 2 more than a multiple of 3 (like five, for example), you only need a 1 or a 4

As you must have noticed, there are 2 numbers to choose from, depending on your case, so therefore, it becomes 216 x 2 = 432 :D

El problema solo está en el primer dado: 6 posibilidades en los tres primeros son 6 x 6 x 6 = 216 posibilidades. Para el ultimo dado solo hay 2: (3,6) o (2,5) o (1,4) por lo tanto hay: 216 x 2 = 432.

Define $f(n)$ as the number of ways to roll $n$ dice and 3 divides $\sum_{i=1}^{n} d_{i}$ . Define $g(n)$ as the number of ways to roll $n$ dice and $\sum_{i=1}^{n} d_{i} \equiv 1 \pmod{3}$ . Define $h(n)$ as the number of ways to roll $n$ dice and $\sum_{i=1}^{n} d_{i} \equiv 2 \pmod{3}$ .

The first thing to note is that $f(1) = g(1) = h(1) = 2$

Now we seek a recursion. Consider $f(n+1)$ . We can have:

$d_{n+1} =3,6$ in $f(n)$ ways each

$d_{n+1} = 1,4$ in $h(n)$ ways each

$d_{n+1} = 2,5$ in $g(n)$ ways each

Therefore we get the recurrence: $f(n+1) = 2(f(n) + g(n) + h(n))$

We can do a similar thing with $g(n)$ and $h(n)$ to get the recurrences: $g(n+1) = 2(f(n) + g(n) + h(n))$ $h(n+1) = 2(f(n) + g(n) + h(n))$

Therefore in fact $f(n+1) = g(n+1) = h(n+1)$ , so $f(n+1) = 6f(n)$ .

Therefore $f(4) = 6^3\times2 = \fbox{423}$

For a harder version of this question, see https://brilliant.org/community-problem/dice-sums-2/?group=1cqtG7Evvhv9

Total number of combinations (without any special case ) would be 6^4 = 1296.Out of these there are equal number of chances of the sum to be -
a) exactly divisible by 3
b)remainder = 1 when divided by 3 c)remainder = 2 when divided by 3
So of the total cases, one third would be the solution = 1296/3 = 432