## An Advent puzzle

By David Shaw, 17 December 2013

See if you can work out this puzzle. You might like to use an advent calendar to solve it, or maybe you could work it out in your head first!

### You will need

On the first day, all the doors were opened.

- Advent calendar (a calendar with 24 or 25 numbered doors on it)

### What to do

In the lead-up to Christmas, Jenny’s parents gave her an advent calendar. But instead of opening one door each day, she had a different way of marking time.

- On 1 December, she opened every door.
This Advent calendar has all the doors in order. If your calendar has the dates all mixed up, you won’t get the open and closed doors lining up like this.

- On 2 December, she went to every second door (starting at 2) and closed each one that was open. After this, all the odd doors were open and all the even doors were closed.
- On 3 December, she went to every third door (starting at 3), opened each one that was closed, and closed each one that was open.
- On 4 December, she went to every fourth door (starting at 4), opened each one that was closed, and closed each one that was open.
- Each day, she repeated this pattern of opening and closing doors. Which doors were open on Christmas day (25 December)?

### What’s happening?

Which doors are open on Christmas day?

There’s quite a bit going on in this puzzle, so it might help to try the first few days and look for patterns. The first pattern you might notice is that after a while you stop opening and closing the small numbered doors. For example, after 4 December, you don’t touch the fourth door, and after 7 December you stop opening and closing door seven.

You might also notice that you open or close a door whenever the date divides the door number exactly. Dates that divide the door number are called factors.

Door six has four factors: 1, 2, 3 and 6. You can arrange them in pairs that multiply together to make six: 2 x 3 = 6 and 1 x 6 = 6. Jenny will go to door six an even number of times – open, close, open, close – and door six will be closed on Christmas day.

Door four has only three factors: 1, 2 and 4. Jenny will go to that door an odd number of times – open, close, open – and it will be open on Christmas day. To see why four has an odd number of factors, we can try to pair them up. 1 and 4 make a pair: 1 x 4 = 4. But 2 pairs with itself: 2 x 2 = 4.

We can look for other doors with self-pairing factors. For example, 3 x 3 = 9, so door nine has a self-pairing factor. So does door sixteen and door twenty-five, if your calendar has one. Door one also has a self-paring factor because 1 x 1 = 1. All these doors have an odd number of factors and will remain open on Christmas day. Numbers with self-paring factors are called ‘square numbers’.

*If you’re after more maths activities for kids, subscribe to Double Helix magazine!*

## 0 comments