$[$ The Pigeonhole Principle (continued) ]
$[\quad$ Estimation + example $\quad]$

The hostess baked a pie for her guests. She may have either 10 or 11 guests. Into what minimum number of pieces should she cut the pie in advance so that it can be evenly divided among either 10 or 11 guests?
If 10 guests arrive, each should receive no less than two pieces.

## Solution

If 10 guests arrive, each should receive no less than two pieces. Indeed, otherwise one of the 10 guests would receive one piece, which is $1 / 10$ of the pie, and if 11 guests arrived, this piece would need to be further divided. Thus, the number of pieces is no less than $2 \cdot 10 = 20$.

We will show that twenty pieces are sufficient. Cut the pie into 10 pieces, each being $\frac{1}{11}$ of the pie, and into 10 pieces, each being $1 / 110$ of the pie. If 10 guests arrive, give each one a large piece and one small piece, totaling $\frac{1}{11} + 1 / 110 = 1 / 10$. If 11 guests arrive, give ten of them one large piece each, which is $1 / 11$, and one guest will receive 10 small pieces.

## Answer

Into 20 pieces.