29. There are two fair dice and their sides are positive integers $a_{1}, \ldots, a_{6}$ and $b_{1}, \ldots, b_{6}$, respectively. After throwing them, the probability of getting a sum of $2,3,4, \ldots, 12$ respectively is the same as that of throwing two normal fair dice. Suppose that $a_{1}+\cdots+a_{6}<b_{1}+\cdots+b_{6}$. What is $a_{1}+\cdots+a_{6}$ ?
29. Answer: 15 .

Let $P(x)=x^{a_{1}}+\cdots+x^{a_{0}}$ and $Q(x)=x^{b_{1}}+\cdots+x^{b_{0}}$. Then
$$
P(x) Q(x)=\left(x+x^{2}+\cdots+x^{6}\right)^{2}=x^{2}(1+x)^{2}\left(1+x+x^{2}\right)^{2}\left(1-x+x^{2}\right)^{2} .
$$

Note that $P(0)=Q(0)=0$ and $P(1)=Q(1)=6, x(1+x)\left(1+x+x^{2}\right)$ is a common divisor of $P(x)$ and $Q(x)$.

Since they are not normal dice,
$$
\begin{array}{l}
P(x)=x(1+x)\left(1+x+x^{2}\right)=x+2 x^{2}+2 x^{3}+x^{4}, \\
Q(x)=x(1+x)\left(1+x+x^{2}\right)\left(1-x+x^{2}\right)=x+x^{3}+x^{4}+x^{5}+x^{6}+x^{8} .
\end{array}
$$

So the numbers of the first dice are $1,2,2,3,3,4$ and that of the second dice are 1,3,4,5,6,8 . Then $a_{1}+\cdots+a_{6}=1+2+2+3+3+4=15$.