Inverse Doomsday

[Albert Chan]

We can also do smarter brute-force.
7 / (1 + 1/4) = 5.6. Instead of trying 1 year at a time, we can jump ahead

lua> f = fn'y: (y+floor(y/4)) % 7'
lua> f(24)
2
lua> f(24-5)
2
lua> f(19-5)
3
lua> f(19-6)
2
lua> f(13-5)
3
lua> f(13-6)
1
lua> f(7-5)
2

f(y) has period of 4*7 = 28 --> f(y)=value has 1/7 = 4/28 solutions

f(y)=2 --> y ≡ {24, 24-5, 19-6, 13-11} ≡ {24, 19, 13, 02} (mod 28)