 ## Mixtures Published on Saturday, 30th August 2014, 07:00 pm; Solved by 148;
Difficulty rating: 100%

### Problem 478

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, cn 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.