Festive Math Puzzles: Solutions and Explanations

Here are the answers to the festive maths quiz I set on December 23. I hope you enjoyed it.

Puzzle 1: You are given nine gold coins that look identical. You are told that one of them is fake, and that this coin weighs less than the real ones. Using a set of old-fashioned balance scales, what is the smallest number of weighings you need to determine which is the fake coin?

Solution: You can do this in just two weighings:

(1) Divide the nine coins into three sets of three, and choose two of these sets to weigh against one another. If one set is lighter than the other, then the fake is one of these three coins. If the two sets weigh the same, then the fake is in the three unweighed coins.

(2) Now take the set with the fake coin, and weigh two of its coins against each other. If one is lighter, that is the fake. If they weigh the same, then the fake is the third coin.

Puzzle 2: You've been transported back in time to help cook Christmas dinner. Your job is to bake the Christmas pie, but all you've got is two egg-timers: one that times exactly four minutes, and one that times exactly seven minutes. How can you time ten minutes exactly?

Solution: There are multiple answers to this puzzle, but supposing the chef wants you cook this pie as quickly as possible, here's how to do it:

- Start both timers at the same time.

- Once the four-minute timer has finished, the seven-minute timer will have three minutes to go. At this point, put the pie in the oven.

- Once the remaining three minutes on the seven-minute timer has finished, turn the seven-minute timer over.

- Let the seven-minute timer run its full course, then take the pie out immediately. The pie will have been in the oven for ten minutes exactly.

Puzzle 3: You are now entrusted with allocating the mulled wine, which is currently in two full ten-litre barrels. The chef hands you one five-litre bottle and one four-litre bottle, both empty. He orders you to fill the bottles with exactly three litres of wine each, without wasting a drop. How can you do this?

Solution: Here is a solution in 11 steps (see table below), recording the quantities of mulled wine in each barrel and bottle. B1 and B2 are the two ten-litre barrels; b5 and b4 are the five-litre and four-litre bottles respectively.

Note: You might have found a quicker solution than mine, but this is what I could come up with!

Puzzle 4: Suppose there are 100 days of Christmas. On the n-th day, you receive £n as a gift, ranging from £1 on the first day to £100 on the final day. Can you calculate the total amount of money you are given, without laboriously adding all 100 numbers together?

Solution: When Carl Friedrich Gauss was posed this question by his maths teacher, the budding mathematician is said to have performed the following calculation:

Let s be the sum of the first 100 digits. Then we can write: s = 1 + 2 + 3 + 4 + … + 99 + 100

But we can also write this backwards: s = 100 + 99 + 98 + … + 4 + 3 + 2 + 1

If we now add these two equations vertically term by term, we see that the left hand side is s + s = 2s.

On the right-hand side, adding vertically again, the sum of every two terms is always the same, namely 101 (1 + 100, 2 + 99 and so on). And there are 100 terms in all - so the easy calculation for the total on the right-hand side is 100 * 101 = 10,100.

Therefore: 2s = 10,100, and s = 5,050. The total amount of money you are given is £5,050.

Puzzle 5: Here's a Christmassy sequence of numbers. The first six in the sequence are: 9, 11, 10, 12, 9, 5 … (Note: the fifth number is 11 in some versions of this puzzle.) What is the next number in this sequence?

Solution: This sequence is the number of letters in each consecutive present given over the 12 days of Christmas. So the answer is 5, for swans. Here's the full list:

Partridge (9), turtle doves (11), French hens (10), calling birds (12), gold rings (9, or 11 for those who sing "golden"), geese (5), swans (5), maids (5), ladies (6), lords (5), pipers (6), drummers (8).

Note: this might seem like a non-mathematical puzzle, but maths - and more broadly, critical and creative thinking - in part relies on spotting patterns that might look a little tenuous at first. Recruitment to the Allies' code-breaking headquarters Bletchley Park during the second world war was partly based on the ability to solve a cryptic crossword .

Puzzle 6: Which of the following 100 statements is the only true one?

• Exactly one statement in this list is false.

• Exactly two statements in this list are false.

  … and so on until:

• Exactly 99 statements in this list are false.

• Exactly 100 statements in this list are false.

Solution: Only the 99th statement in this list is true. Since there are 100 statements, and the n-th statement asserts that exactly n statements in the list are false, this can only be true when n = 99.

Puzzle 7: You and your friends Arthur and Bob are wearing Christmas hats that are either red or green. Nobody can see their own hat but you can all see the other two. Arthur's and Bob's hats are both red.

You are all told that at least one of the hats is red. Arthur says: "I do not know what colour my hat is." Then Bob says: "I do not know what colour my hat is." Assuming your friends have impeccable logic, can you deduce what colour your Christmas hat is?

Solution: Your hat must be red. If your hat were green, then both Arthur and Bob would see one green and one red hat. So when Arthur says that he doesn't know the colour of his hat, Bob could immediately deduce that his hat was red. But since Bob doesn't know the colour of his hat, Bob must be seeing two red hats, and so you can deduce that your hat is red.

Puzzle 8: There are three boxes under your Christmas tree. One contains two small presents, one contains two pieces of coal, and one contains a small present and a piece of coal. Each box has a label on it that shows what's inside - but the labels have got mixed up, so every box currently has the wrong label on it.

You are told that you can reach in and take out one object from just one box. Which box should you choose in order to then be able to switch the labels so that every label correctly corresponds to the contents of its box?

Solution: Since all the boxes have the wrong labels, you know that if you open the box currently labelled as containing one small present and one piece of coal, you will either see two small presents or two pieces of coal.

Suppose you open it and see two small presents. Then the label of two small presents must be fixed to this box. And since you also know that every box originally had the wrong label, the label of one small present and one piece of coal should go on the box currently labelled two pieces of coal. Finally, the two pieces of coal label belongs to the box originally labelled two small presents.

Puzzle 9: There is a one-litre bottle of orange juice and a one-litre bottle of apple juice in the kitchen. Jack puts a tablespoon of orange juice into the bottle of apple juice, then stirs it around so it's evenly mixed. Now Jill takes a tablespoon of liquid from that apple juice bottle puts it back in the bottle of orange juice. Is there now more orange juice in the bottle of apple juice, or more apple juice in the bottle of orange juice?

Solution: They are the same. This is a nice example of "invariance" - a term that comes up a lot in mathematics.

After all the adding of tablespoons of juice and all the mixing, the amount of orange juice in the apple juice bottle must have replaced the same amount of apple juice that was originally in the apple juice bottle, because the amount of liquid in each bottle is still one litre (they have remained invariant).

This explanation can feel unsatisfactory when you first read it. But exploiting the power of invariance allows you to deduce that the amounts must be the same, without any calculation.

Puzzle 10: In Santa's home town, all banknotes carry pictures of either Santa or Mrs Claus on one side, and pictures of either a present or reindeer on the other. A young elf places four notes on a table showing the following pictures in this order:

Now an older, wiser elf tells him: "If Santa is on one side of the note, a present must be on the other." Which notes must the young elf turn over to confirm what the older elf says is true?

Solution: First, the young elf should turn over the banknote with Santa on it. If there isn't a present on the other side, then the older elf is lying. Next, the young elf should turn over the reindeer banknote to confirm that Santa is not on the other side. Again, if Santa were on the other side, the older elf would be lying.

It might be tempting to turn over the present banknote. But the older elf only says "if Santa, then present", which doesn't imply "if present, then Santa". So it doesn't matter whether Santa or Mrs Claus is on the other side of the present banknote - and it also doesn't matter what is on the other side of the Mrs Claus banknote, because the older elf doesn't say anything about those notes.

Bonus puzzle solution

Santa travels on his sleigh from Greenland to the North Pole at a speed of 30 miles per hour, then immediately returns from the North Pole to Greenland at a speed of 40 miles per hour. What is the average speed of Santa's entire journey?_

Solution: This puzzle is perhaps an example of what psychologist Daniel Kahneman called Thinking Fast and Slow. Our fast-thinking system might say "just take the average", and so we would guess 35 miles per hour. A reasonable answer, but wrong.

Our slower-thinking system - which is effortful to use, requiring tools like algebra and critical thinking - is needed here. First, let's set up some variables:

- let d be the distance from Greenland to the North Pole.

- let t₁ be the time taken on the out-journey.

- let t₂ be the time taken on the return journey.

Using the standard equation "speed = distance divided by time", we can say:

30 = d/t₁ and 40 = d/t₂

Rearranging these equations, we also know that t₁ = d/30 and t₂ = d/40

Since Santa travels the same distance there and back, his total distance travelled is 2d. And the average speed for the total journey is this total distance divided by the total time taken: 2d/(t₁ + t₂)

Using all of the above, we can say the average speed of Santa's journey is 2d/(d/30 + d/40)

Now, (d/30 + d/40) = (4d/120 + 3d/120) = 7d/120

So Santa's average speed = 2d / (7d/120) = 240/7 = 34.3

In this equation, the 'd's cancel out. This means we can work out the average speed of the journey without knowing either the distance or the time it took Santa to make his journey. This is the power of algebra: it allows you to use and manipulate quantities, using them as place holders even when you don't know what the quantities are.

The answer is that Santa travelled at an average speed of 34.3 miles per hour.

Neil Saunders does not work for, consult, own shares in or receive funding from any company or organisation that would benefit from this article, and has disclosed no relevant affiliations beyond their academic appointment.