Function aoc::year2019::day16::binomial_mod_10

fn binomial_mod_10(n: usize, k: usize) -> usize

Computes C(n, k) % 10

Solving the Chinese remainder theorem for the special case of two congruences:

    x ​≡ a₁ (mod n₁) ​≡ a₁ (mod 2)
    x ​≡ a₂ (mod n₂) ≡ a₂ (mod 5)
    N = n₁n₂ = 10
    y₁ = N / n₁ = 5
    y₂ = N / n₂ = 2
    z₁ = y₁⁻¹ mod n₁ = 5⁻¹ mod 2 = 1
    z₂ = y₂⁻¹ mod n₂ = 2⁻¹ mod 5 = 3
    x ≡ a₁y₁z₁ + a₂y₂z₂ (mod 10) ≡ 5a₁ + 6a₂ (mod 10)