Toasting Three at a Time

Lower bound. Four slices have $8$ sides to toast. The pan toasts at most $3$ sides per minute, so at least $\lceil 8/3\rceil = 3$ minutes are needed.

Construction. Call the slices $A,B,C,D$ and their sides $A_1,A_2,\dots,D_2$. Toast:

• minute 1: $A_1, B_1, C_1$;
• minute 2: $A_2, B_2, D_1$;
• minute 3: $C_2, D_2$.

In every minute the sides belong to different slices, and each side is toasted exactly once. So all four slices are done after $3$ minutes. The key is that slice $C$ is taken off after its first side and finished later. This keeps the pan full in the first two minutes.

Generalisation. Let a pan hold $k$ slices, and let there be $n$ slices to toast. The minimum time is $\max\bigl(2,\lceil 2n/k\rceil\bigr)$.

• Case $n<k$. Two minutes suffice and are needed, since each slice needs two separate minutes.
• Case $n\ge k$. Write the $2n$ sides in the order $1_1,2_1,\dots,n_1,1_2,2_2,\dots,n_2$. Toast them $k$ at a time in this order. The two sides of a slice are $n\ge k$ places apart, so they never fall in the same block. This takes $\lceil 2n/k\rceil$ minutes, which matches the counting bound.

On its own, the counting bound $\lceil 2n/k\rceil$ can fail when there are fewer slices than places, e.g. $2$ slices in a $5$-slice pan.