# Problem 478: Mixtures
Let us consider mixtures of three substances: A, B and C. A mixture can
be described by a ratio of the amounts of A, B, and C in it, i.e.,
(a : b : c). For example, a mixture described by the ratio (2 : 3 : 5)
contains 20% A, 30% B and 50% C. For the purposes of this problem, we
cannot separate the individual components from a mixture. However, we
can combine different amounts of different mixtures to form mixtures
with new ratios. For example, say we have three mixtures with ratios
(3 : 0 : 2), (3 : 6 : 11) and (3 : 3 : 4). By mixing 10 units of the
first, 20 units of the second and 30 units of the third, we get a new
mixture with ratio (6 : 5 : 9), since:
(10·3/5 + 20·3/20 + 30·3/10 : 10·0/5 + 20·6/20 + 30·3/10 : 10·2/5 + 20·11/20 + 30·4/10)
= (18 : 15 : 27) = (6 : 5 : 9) However, with the same three mixtures, it
is impossible to form the ratio (3 : 2 : 1), since the amount of B is
always less than the amount of C. Let n be a positive integer. Suppose
that for every triple of integers (a, b, c) with 0 ≤ a, b, c ≤ n and
gcd(a, b, c) = 1, we have a mixture with ratio (a : b : c). Let M(n) be
the set of all such mixtures. For example, M(2) contains the 19 mixtures
with the following ratios: {(0 : 0 : 1), (0 : 1 : 0), (0 : 1 : 1),
(0 : 1 : 2), (0 : 2 : 1), (1 : 0 : 0), (1 : 0 : 1), (1 : 0 : 2),
(1 : 1 : 0), (1 : 1 : 1), (1 : 1 : 2), (1 : 2 : 0), (1 : 2 : 1),
(1 : 2 : 2), (2 : 0 : 1), (2 : 1 : 0), (2 : 1 : 1), (2 : 1 : 2),
(2 : 2 : 1)}. Let E(n) be the number of subsets of M(n) which can
produce the mixture with ratio (1 : 1 : 1), i.e., the mixture with equal
parts A, B and C. We can verify that E(1) = 103, E(2) = 520447,
E(10) mod 118 = 82608406 and E(500) mod 118 = 13801403. Find
E(10 000 000) mod 118.