# Ah...finally, a solution.

**Question:** When hosting a party, I put out 43 glasses of champagne. At the end of the night all glasses are empty. I deduce that someone has had at least three glasses and shouldn’t be driving home.

How many people (at most) came to my party?

### Answer: **21**

**Short solution: **

Suppose 21 people come to my party and I try to distribute the glasses as evenly as possible amongst my guests. I start by giving everyone two glasses each. This uses up 42 of the glasses but there is still one left over. This glass must be given to someone who will then have three glasses! A little thought should convince you that no matter how I share out the glasses, someone will always have to be holding three. Now suppose that 22 people came. If 21 of them had two glasses and the remaining person just had one glass then all the champagne will have been drunk and no one has had more than 2 glasses. Therefore, for 22 people, it's quite possible that no one had three glasses. So, in order for me to make my deduction that someone has had three glasses or more, there must have only been at most 21 people at the party.

**Detailed solution:**

There are three essential pieces of information in this question which we must identify to get to our solution.

- There are 43 glasses of champagne. (...obviously).
- I can deduce that at least someone has had three glasses or more.
- We are asked for the
*maximum*number of people that came (such that this deduction is possible).

This last point gives a great deal away about the nature of the problem. Consider the situation in which 2 people came to my party. Clearly one of them at least has had more than two glasses. For comparison, consider if 43 people had turned up. It's quite possible that each of them had, for example, one glass each. Therefore, for 43 people, we cannot deduce for certain that someone must have had at least three glasses. This 'playing' with different numbers might seem quite a roundabout way of approaching the problem but it has actually been very useful because it gives us intuition about the problem and tells us that the number we are looking for must be somewhere between 2 and 43. At this point it would be very useful to use the following idea, commonly referred to as 'The Pigeonhole Principle' which essentially says the following.

'If I have more pigeons than pigeonholes then at least one hole must contain more than one pigeon.'

This might seem like a fairly straightforward idea, or even common sense, but this is an incredibly powerful tool that is used in all areas of maths. Whilst the principle is simple, the real trick is in realising when it would be useful to use it and then working out what the 'pigeons' and 'pigeonholes' should be. In our problem, the champagne glasses are our pigeons and the guests at the party are the pigeonholes. Imagine each guest (or pigeonhole) is able to happily have at most two glasses (or pigeons), then we need to work out how many guests will just tip the balance so that just one person will have to have three. Therefore we can assume just one person has exactly three glasses, leaving 40 which can perfectly 'fill' 20 pigeonholes (guests with two glasses). This gives us a maximum of 21 guests for which someone has to have at least three glasses. We can always check that this is the maximum number by showing (as above) that 22 people could finish off the champagne without having to have more than two glasses each.

Disclaimer: I'm pretty sure two glasses is too many to be driving home on anyway so my irresponsibly drinking guests were purely imagined for the construction of the question!