<s>[INST] 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?
 [/INST] Let's use Zhuli's notation.
$c$ has a probability of $\frac{1}{2}$ to be even.
Now if $\frac{c}{2}$ is even (i.e., $4|c$), then $4|2b+c \iff b$ is even. If $\frac{c}{2}$ is odd, then $4|2b+c \iff b$ is odd. Both conditional probabilities are $\frac{1}{2}$ therefore the marginal probability $4|2b+c$ is $\frac{1}{4}$.
Next we apply the same reasoning on $2a + (b+\frac{c}{2})$ we get the marginal probability that $4 | 2a + (b+\frac{c}{2})$, or equivalently, $8 | 4a + 2b +c$, is $\frac{1}{8}$. Q.E.D.
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Alternatively we could avoid conditional probabilities and just use independence and marginal probabilities.
First note that $\Pr(2|a)=\Pr(2|b)=\Pr(2|c)=\frac 12$ and $a,b,c$ are independent.
Next,
$$
\Pr(4|2b+c)=\Pr(2|c,4|c,2|b) + \Pr(2|c, 4\nmid c, 2\nmid b)
$$
$$
=\Pr(2|c, 4|c) \Pr(2|b)+\Pr(2|c, 4\nmid c)\Pr(2\nmid b) \tag{independence}\\
= \frac 12 \left( \Pr(2|c, 4|c) + \Pr (2|c, 4\nmid c)\right) = \frac 12 \Pr(2|c)=\frac 14.
$$
Finally,
$$
\Pr(8|4a+2b+c)=\Pr(4|2b+c,8|2b+c,2|a) + \Pr(4|2b+c, 8\nmid 2b+c, 2\nmid a)\\
=\Pr(4|2b+c,8|2b+c)\Pr(2|a) + \Pr(4|2b+c, 8\nmid 2b+c)\Pr(2\nmid a)\\
=\frac 12 \left(\Pr(4|2b+c,8|2b+c) + \Pr(4|2b+c, 8\nmid 2b+c)\right)\\
=\frac 12 \Pr(4|2b+c)=\frac 12 \cdot \frac 14 = \frac 18.\blacksquare
$$
Remark: We can prove, via induction, the generalization proposed by Anant using the technique above.
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