Dice Dicey Dice Dice 2

Suppose that $K$ dice are used, with sides $N_1,N_2,\ldots,N_K$ . If the probability generating function of the $j$ th dice is $G_j(t)$ , then $G_j(t) \; = \; \frac{1}{N_j}P_j(t)$ where $P_j(t)$ is a polynomial with nonnegative integer coefficients such that $P_j(1) = N_j$ for each $j$ . In addition we must have $\prod_{j=1}^K G_j(t) \; =\; \frac{1}{60}\sum_{j=1}^{60} t^j$ so that the sum of the throws of all the dice gives a uniform distribution on $\{1,2,3,\ldots,60\}$ . Thus $60P_1(t)P_2(t) \cdots P_K(t) \; = \; N_1N_2 \cdots N_k \big(t + t^2 + \cdots + t^{60}\big)$ and so $t$ must divide one of the polynomials $P_j(t)$ ,. Without loss of generality we assume that $P_1(t) = t\hat{P}_1(t)$ for some polynomial $\hat{P}_1(t)$ , and hence $\begin{aligned} 60\hat{P}_1(t) P_2(t) \cdots P_K(t) & = N_1N_2\cdots N_k(1 + t +t^2 + \cdots t^{59}) \\ 60\hat{P}_1(0)P_2(0) \cdots P_K(0) & = N_1N_2 \cdots N_k \end{aligned}$ and so at least one of $N_1, N_2, \ldots, N_K$ must be divisible by $3$ , and at least one must be divisible by $5$ . Thus one of the dice must be a $20$ -gon, and one of the other dice must be either a $6$ -gon or a $12$ -gon.

Thus the least possible number of faces used will be $26$ if we can achieve what we want using just a $6$ -gon and a $20$ -gon. But since $(t + t^2 +t^3 + t^4 + t^5 + t^6)(1 + t^6 + t^{12} + t^{18} + t^{24} + t^{30} + t^{36} + t^{42} + t^{48} + t^{54}) = t + t^2 + t^3 + \cdots t^{60}$ we see that $\frac{1}{60}\big(t + t^2 + \cdots + t^{60}\big) \; = \; P_1(t) P_2(t)$ where $\begin{aligned} P_1(t) & = \frac16\big(t + t^2 + t^3 + t^4 + t^5 + t^6\big) \\ P_2(t) & = \frac{1}{10}\big(1 + t^6 + t^{12} + t^{18} + t^{24} + t^{30} + t^{36} + t^{42} + t^{48} + t^{54}\big) \\ & = \frac{1}{20}\big(2 + 2t^6 + 2t^{12} + 2t^{18} + 2t^{24} + 2t^{30} + 2t^{36} + 2t^{42} + 2t^{48} + 2t^{54}\big) \end{aligned}$ and so we can obtain the desired result by combining a normal $6$ -sided die labelled with $\{1,2,3,4,5,6\}$ and a $20$ -sided die labelled with $\{0,0,6,6,12,12,18,18,24,24,30,30,36,36,42,42,48,48,54,54\}$ .

The answer is $\boxed{26}$ .