Reply To: Puzzles

[upsidedown]

@F6EXB_the_frenchy: it’s on the first page

@ByteInBits: answer for #107332

It depends whether the contestants round the 1/5 that’s put in the hole up or down.

1. If they round up, 3128
2. If they round down, 3122

We’re after the smallest number of coins, so I’ll say 3122.

Supposing the coins were a continuous quantity, I have a closed form for the initial number of coins c₀ in terms of: n, the final number of coins per player; p+1, the total number of players; and 1-r, the ratio which is put in the hole.

The minimum number of coins before n of the players went into the room together was np + 1

I’m now only going to consider the iterative process of each player going into the room alone. If the number of coins before going in the room was cᵢ, then cᵢ₊₁ is defined by:

cᵢ₊₁ = r(cᵢ + 1)

The closed form for cᵢ is:

cᵢ = rⁱc₀ – (1 – rⁱ)r / (1 – r)

So if cₚ = np + 1,

rᵖc₀ – (1 – rᵖ)r / (1 – r) = np + 1

c₀ = (np + 1 + (1 – rᵖ)r / (1 – r)) / rᵖ

…although this is useless because the coins are not, in fact, a continuous quantity 🙂