Twelve days of Christmas

As our house is getting rather full of drumming drummers and maids-a-milking, my true love decided on a different gifting strategy this year. I will receive a number of gifts based on the formula

    \[g=\left[ \sqrt{2n}+1/2 \right]\]

where g is the number of gifts, n is the day of Christmas (e.g. n=2 for the second day of Christmas) and [x] denotes the greatest integer less than or equal to x. Our regular gift stockist only has the usual Christmas fayre for sale so my true love has decided that

  • when g=1 I will receive 1 partridge complete with pear tree
  • when g=2 I will receive 2 turtle doves
  • when g=3 I will receive 3 French hens

and so on (following items from previous years).

How many “calling birds” (g=4) will I receive?

How many “swimming swans” (g=7) will we have to accommodate?

What will I receive on the 12th day of Christmas?

How many gifts will I receive overall this year?

Reveal Solution
Calculating a few terms gives a good feel for things
n=1 g=\left[\sqrt{2}+1/2 \right] since 1<\sqrt{2}<1.5 this gives 1.5<\sqrt{2}+1/2<1.5 so g=1.

n=2 g=\left[ \sqrt{4}+1/2 \right] so g=2.

n=3 g=\left[ \sqrt{6}+1/2 \right] since 2<\sqrt{6}<2.5 this gives 2.5<\sqrt{6}+1/2<3 so g=2.

n=4 g=\left[ \sqrt{8}+1/2 \right] since 2.5<\sqrt{8}<3 this gives 3<\sqrt{8}+1/2<3.5 so g=3.

The next change will take place when \sqrt{2n}>3.5 i.e. 2n>12.25 so n>6.125 so, on the 7th day g=4 and on days 5 and 6 g=3.

To go to g=5, \sqrt{2n}>4.5 i.e. 2n>20.25 so n>10.125 . This is day 11.

The sequence would look like this: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5.

There are 4 days on which I would receive 4 calling birds so I would have 16 calling birds in total.

Fortunately, no swimming swans need to be accommodated since g<7 for all 12 days.

On the 12th day of Christmas I will receive 5 gold rings.

So this year I would receive 1+4+9+16+10=40 gifts:

  • 1 partridge complete with pear tree
  • 4 turtle doves – only a small dovecote needed for these
  • 9 French hens – or “poules françaises” as we like to call them
  • 16 calling birds – a tad noisy but never mind
  • 10 gold rings

Not a bad haul and much more manageable than previous years!

Problem Page Coordinator: Claire Baldwin – Mathematics in Education and Industry

Acknowledgement: The IMA are indebted to MEI for sourcing and supplying Mathematics Today with these well-known puzzles.

First published in Mathematics Today (December 2017)

Image credit: Presents by Alice Harold / Flickr / CC BY-NC-ND 2.0

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