Question 37885
If we have 1200 coins on day zero and we remove 20% on day one, then 80% remains...
D0 = 1200
D1 = 1200(.8)
D2 = 1200(.8)(.8)
D3 = 1200(.8)(.8)(.8)
.
.
.
D7 = 1200(.8)^7 = 251+ coins

This is not a great problem because of the fractions of coins, but that is about it...