Martin Gardner was a journalist, a novelist, a magician, a philosopher and one of the earliest public debunkers of pseudoscience.

Yet he was probably best known – and most loved – for popularising mathematical puzzles.

In his monthly column Mathematical Games, which he wrote in Scientific American between the 1950s and the 1980s, he introduced many brainteasers as well as giving old classics new twists.

“Puzzles can lead you into almost every branch of mathematics.”- Martin Gardner

Gardner wrote dozens of books on puzzles and recreational maths – here are eight puzzles taken from them. Can you solve them?

## 1. Crazy cut

You are to make one cut (or draw one line) – of course it needn’t be straight – that will divide the figure into two identical parts.

## 2. The coloured socks

Ten red socks and ten blue socks are all mixed up in a dresser drawer. The 20 socks are exactly alike except for their colour. The room is in pitch darkness and you want two matching socks. What is the smallest number of socks you must take out of the drawer in order to be certain that you have a pair that match?

## 3. Twiddled bolts

## 4. The fork in the road

## 5. Three squares

Using only elementary geometry (not even trigonometry), prove that angle C equals the sum of angles A and B.

## 6. Cutting the pie

## 7. The mutilated chessboard

The props for this problem are a chessboard and 32 dominoes. Each domino is of such size that it exactly covers two adjacent squares on the board. The 32 dominoes therefore can cover all 64 of the chessboard squares. But now suppose we cut off two squares at diagonally opposite corners of the board and discard one of the dominoes.

Is it possible to place the 31 dominoes on the board so that all the remaining 62 squares are covered? If so, show how it can be done. If not, prove it impossible.

## 8. The two spirals

__Scroll down to see the solutions to the puzzles__

## 1. Crazy cut

Solution:

The hint wasn’t a red herring. The line isn’t straight.

## 2. The coloured socks

Solution: Three socks.

With two socks it is possible to have one red and one blue. But with three there is always a matching pair since either you will have chosen three of the same colour, or a matching pair and an odd one out.

## 3. Twiddled bolts

Solution: (c). The heads of the twiddled bolts move neither inward nor outward. The movements cancel each other out – like a person walking up an escalator at the same rate that it is moving down. If you get two bolts, or screws, and try it, it’s fun to see.

## 4. The fork in the road

Solution: The challenge here is to find a question that forces a liar to lie about a lie and hence tell the truth.

This works: point at one of the forks and ask the native: “If I were to ask you if this road leads to the village, would you say yes?”

If the fork is the correct one, a liar would answer no to the direct question ‘does this road lead to the village?’ and thus his (lying) answer to the actual question must be yes.

The truth-teller will also answer yes if the road is the correct one.

## 5. Three squares

Using only elementary geometry (not even trigonometry), prove that angle C equals the sum of angles A and B.

Solution: Here’s one way to do it.

Construct the additional squares indicated by dotted lines. Angle C is the same as the sum of angles A and D, since they are both made by cutting a square in half along the diagonal. Angle B equals angle D because they are corresponding angles of similar right triangles (made by cutting a rectangle made from two squares in half along the diagonal). That means B can be substituted for D, which automatically makes the C equals the sum of A and B.

## 6. Cutting the pie

Solution: You can do this by trial and error, but the sketch will get a bit messy, and it is not particularly insightful. Better to think up a rule. So, think about what is happening when you add straight cuts.

The first cut splits the pie into 2 pieces.

The second cut makes 2 more pieces, bringing the total to 4

The third cut makes 3 more pieces, bringing the total to 7.

It looks like each cut adds the total number of pieces by the number of the cut – and with a little more thought we can see why this is true. I’ll leave that for homework…or click here.

So, the fourth cut will make as many as 4 new pieces, the fifth cut 5 and the sixth cut 6. The largest number of pieces therefore is 22.

## 7. The mutilated chessboard

Solution: The chessboard without two opposite corner squares cannot be covered with 31 dominoes.

First, we need to realise that a domino always covers two adjacent squares of the chessboard, which must always have the opposite colour.

If you were able to cover all but two squares of the mutilated chessboard, the remaining two squares would be of the same colour (the opposite colour to the removed corners). Since adjacent squares are of opposite colour, the remaining squares are not adjacent and hence cannot be covered by the last domino.

## 8. The two spirals

Solution: The spiral on the left is the single rope.

Source: The Guardian/Alex Bellos; Illustrations: National Museum of Mathematics ; Thinkfun.com