Probability of number formed from dice rolls being multiple of 8 
A fair 6-sided die is tossed 8 times. The sequence of 8 results is recorded to form an 8-digit number. For example if the tosses give {3, 5, 4, 2, 1, 1, 6, 5}, the resultant number is $35421165$. What is the probability that the number formed is a multiple of 8.

I solved this by listing all possibilities for the last 3 digits that give multiples of 8, and found this to be $\frac{1}{8}$. 
The solution key agrees with my answer, but also says that "There are quicker ways to solve the problem using a more advanced knowledge of number theory"

What would be a faster way to solve this using number theory?

We only care about the last three digits obviously, so the possible sums of just those three die are $100a + 10b + c$, where $a$, $b$, and $c$ are all between $1$ and $6$, inclusive.
We check the divisibility by taking this modulo $8$. If the result is equivalent to $0$, then it is divisible.
Setting up the equivalence condition:
$$
100a+10b+c \equiv 0 \pmod{8} \\
4a + 2b + c \equiv 0 \pmod{8} \\
2(2a+b) + c \equiv 0 \pmod{8} \\
2(2a+b) \equiv -c \pmod{8} \\
2a + b \equiv -\frac{c}{2} \pmod{4}
$$
The range of $2a+b$ is between $3$ and $18$, inclusive. Each of the values can be obtained in three different ways, except $\{3,4,17,18\}$ each can only be obtained one way, and $\{5,6,15,16\}$ each can be done two ways (there are $8$ integers that can be obtained $3$ ways).
Three cases: $-\frac{c}{2}$ must be $3$, $2$, or $1$ modulo $8$. Because the consecutive $8$ integers in the middle can be done $3$ ways, each case has exactly $2$ numbers that satisfy the equivalence, giving us $3\cdot{3}\cdot{2} = 18$ cases. Taking the other integers modulo $4$, only $1$ number from each of the other sets is not congruent to any of $\{3,2,1\}$, giving us $3\times{3} = 9$ more cases, totalling $18+9 = 27$.
Final probability: $\frac{27}{6^3} = \frac{1}{8}$.
