The number of such colorings is $2^{20} 3^{10} = 61917364224$. Identify the three colors red, white, and blue with (in some order) the elements of the field \mathbb{F}_3 of three elements (i.e., the ring of integers mod 3). The set of colorings may then be identified with the \mathbb{F}_3-vector space \mathbb{F}_3^E generated by the set $E$ of edges. Let $F$ be the set of faces, and let \mathbb{F}_3^F$be the \mathbb{F}_3-vector space on the basis$F$; we may then define a linear transformation$T: \mathbb{F}_3^E \to \mathbb{F}_3^F$taking a coloring to the vector whose component corresponding to a given face equals the sum of the three edges of that face. The colorings we wish to count are the ones whose images under$T$consist of vectors with no zero components. We now show that$T$is surjective. (There are many possible approaches to this step; for instance, see the following remark.) Let$\Gamma$be the dual graph of the icosahedron, that is,$\Gamma$has vertex set$F$and two elements of$F$are adjacent in$\Gamma$if they share an edge in the icosahedron. The graph$\Gamma$admits a hamiltonian path, that is, there exists an ordering$f_1,\dots,f_{20}$of the faces such that any two consecutive faces are adjacent in$\Gamma$. For example, such an ordering can be constructed with$f_1,\dots,f_5$being the five faces sharing a vertex of the icosahedron and$f_{16},\dots,f_{20}$being the five faces sharing the antipodal vertex. For$i=1,\dots,19$, let$e_i$be the common edge of$f_i$and$f_{i+1}$; these are obviously all distinct. By prescribing components for$e_1,\dots,e_{19}$in turn and setting the others to zero, we can construct an element of \mathbb{F}_3^E whose image under$T$matches any given vector of \mathbb{F}_3^F$ in the components of $f_1,\dots,f_{19}$. The vectors in \mathbb{F}_3^F$obtained in this way thus form a 19-dimensional subspace; this subspace may also be described as the vectors for which the components of$f_1,\dots,f_{19}$have the same sum as the components of$f_{2},\dots,f_{20}$. By performing a mirror reflection, we can construct a second hamiltonian path$g_1,\dots,g_{20}$with the property that$g_1 = f_1, g_2 = f_5, g_3 = f_4, g_4 = f_3, g_5 = f_2$. Repeating the previous construction, we obtain a \emph{different} 19-dimensional subspace of \mathbb{F}_3^F$ which is contained in the image of $T$. This implies that $T$ is surjective, as asserted earlier. Since $T$ is a surjective homomorphism from a 30-dimensional vector space to a 20-dimensional vector space, it has a 10-dimensional kernel. Each of the $2^{20}$ elements of \mathbb{F}_3^F$with no zero components is then the image of exactly$3^{10}$ colorings of the desired form, yielding the result.