The 12 Days of Christmas Puzzles
Dr Tom Crawford, St Edmund Hall
Puzzle 1:
If I set a puzzle every day of the advent period (1–25 December) and spend 1 minute on the first puzzle, 2 minutes on the second, 3 minutes on the third, and so on, with the final one being 25 minutes on the 25th puzzle, what is the total amount of time I will spend writing puzzles?
Answer:
1 + 2 + 3 + … + 25 = 325. There is a faster way to do this which was first discovered by the mathematician Gauss when he was still at school. If you pair each of the numbers in your sum, eg. 0 + 25, 1 + 24, 2 + 23, etc. up to 12 + 13, then you have 13 pairs which each total 25 and so the overall total is 25*13 = 325. The same method works when adding up the first n numbers, with the total always being n(n+1)/2.
Puzzle 2:
December 6th marked my birthday and to celebrate I travelled to Kiev with 4 friends. If I order a drink on the flight out and then each of my friends orders twice as many as the person before, how many drinks do we order in total?
Answer:
1+2+4+8+16 = 31.
Puzzle 3:
This morning I built a snowman using three spheres of radius 0.5m, 0.4m and 0.2m. However, the sun has since come out and the snowman is starting to melt at a rate of 0.01 m³ per minute. How long will it take for him to disappear completely?
Answer:
Volume of a sphere = (4/3)*π*radius³ and so the total volume of snow = 0.52 + 0.27 + 0.03 = 0.82 m³. Melting at a rate of 0.01 m³ per minute means the snowman will be gone after only 82 minutes!
Puzzle 4:
Suppose a newly-born pair of elves, one male, one female, are living together at the North pole. Elves are able to mate at the age of one month so that at the end of its second month a female elf can produce another pair of offspring. Suppose that the elves never die, and that the female always produces one new pair (one male, one female) every month from the start of the third month on. After one year, how many pairs of elves will there be?
Answer:
This problem is actually a very famous sequence in disguise…
The first new pair is born at the start of the third month giving 2 pairs after three months. The question tells us that we have to wait one whole month before the new offspring can mate and so only the original pair can give birth during the fourth month which leaves a total of 3 pairs after four months. For the fifth month, both the original pair, and the first-born pair can now produce offspring and so we get two more pairs giving a total of 5 after five months. In month six, the second-born pair can now also produce offspring and so in total we have three offspring-producing pairs, giving 8 pairs after 6 months.
At this point, you may have spotted that the numbers follow the Fibonacci sequence, which is created by adding the previous two numbers together to get the next one along. The first twelve numbers in the sequence are below, which gives an answer of 144 — no wonder Santa is able to make so many toys!
Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …
Puzzle 5:
On Christmas day I have 11 people coming to dinner and so I’m working on the seating plan ahead of time. For a round table with exactly 12 chairs, how many different seating plans are possible?
Answer:
I have 12 choices of where to place the first person, 11 for the second, 10 for the third and so on, which gives 12*11*10*9*8*7*6*5*4*3*2*1 = 12! (read as 12 factorial) in total. BUT for any given seating plan we can rotate around the table one place to get the same order, which means we have in fact over counted by a factor of 12. Therefore, the total number is 11! = 39,916,800.
Puzzle 6:
My front yard is covered in snow and I need to clear a path connecting my front door to the pavement and then back to the centre of the garage. If each square in the diagram is 1m x 1m what is the shortest possible path?
Answer:
Reflect the yard in the pavement and draw a straight line connecting the front door to the edge of the garage closest to the front door (blue). Then add the same line from the ‘reflected’ front door at the top back down to the garage at the bottom (orange). The final shortest path is found by combining both paths for a valid one in the original diagram.
The length is found using Pythagoras’ Theorem. From the door to the pavement we have length
and from the pavement to the garage the length is
giving a total length of 2.23 + 3.35 = 5.58m.
Puzzle 7:
The first night of Chanukah is December 22nd when the first candle is lit. If it burns at a rate of 0.05cm per hour, how tall does the candle need to be to last the required 8 days?
Answer:
8 days = 8*24 hours = 192 hours. 192*0.05 = 9.6cm.
Puzzle 8:
If you have a square chimney which is 0.7m across, assuming Santa has a round belly what is the maximum waist size that can fit down the chimney?
Answer:
Chimney diameter = 0.7m so the maximum circumference (or waist size) that will fit is 0.7π = 2.2m or 88 inches!
Puzzle 9:
On Christmas Eve Santa needs to visit each country around the world in 24 hours. Assuming time stands still whilst he is travelling, how long can he spend in each country?
Answer:
Using the UN list of 193 countries, Santa has 24 * 60 = 1440 minutes total, which means spending only 7.5 minutes in each country!
Puzzle 10:
I got carried away with buying presents this year and now have more than can fit into my stocking. If the stocking has a maximum capacity of 150, and my presents have the following sizes: 16, 27, 37, 65, 52, 42, 95, 59; what is the closest I can get to filling the stocking completely?
(NB: I am not looking for the highest number of presents that will fit, but the largest total that is less than or equal to 150).
Answer:
150 exactly with 16 + 27 + 42 + 65 = 150.
Puzzle 11:
Santa has 8 reindeer, and each one can pull a weight of 80kg. If Santa weighs 90kg, his sleigh 180kg, and each present weighs at least 3kg, what is the maximum number of presents that can be carried in a single trip?
Answer:
We have 8 reindeer each with a capacity of 80kg giving a total of 640kg that can be carried. Subtracting the 90kg for Santa and 180kg for the sleigh leaves 370kg available. Dividing this by 3 gives 123.33 so a maximum of 123 presents can be carried at once.
Puzzle 12:
To mark the end of the 12 days of Christmas each student at the University of Oxford has kindly decided to donate some money to a charity of their choice. If the first person donates £12 and everyone after donates exactly half the amount of the person before them (rounding down to the nearest penny), how much will be donated in total?
Answer:
12 + 6 + 3 + 1.5 + 0.75 + 0.37 + 0.18 + 0.09 + 0.04 + 0.02 + 0.01 + 0 + 0 + 0 + …
The donations stop after the 11th person giving a total of £23.87. Even if we had allowed donations of part of a penny the total would never quite reach £24.00. This is an example of an infinite sum (or Geometric Series) where the total is always two times the first number.
Author
Dr Tom Crawford — for more maths fun check out Tom’s website tomrocksmaths.com and follow him on Twitter, Facebook, Instagram and YouTube @tomrocksmaths.
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