In the past years, the world has been gripped by a series of math and logic puzzles that were originally set for children.

Remember the parked car puzzle? Originally set as a test for primary school children in Hong Kong but many adults still found tricky to solve. Then came Cheryl's birthday, set for 15 to 16-year-olds in Singapore but seemingly impossible for anyone to solve. And finally, the rod and string conundrum which stumped 96 per cent of top math students in the US when it first appeared back in the 1990s.

Because all have proved so popular, and many of you seem to have already solved the original three, here are eight more puzzles to leave you chewing your pencil and pulling out your hair.

And before you scroll to the answers in the second part of the article, try to solve the puzzles by yourself.

## 1. How to beat Roger Federer at Wimbledon (via Peter Winkler)

Thanks to a set of temporary magical powers you are in the final of the Wimbledon tennis championships up against seven-time winner Roger Federer. Your powers cannot last for the whole match and you must therefore choose the optimum time for them to run out. What is the score that gives you the maximum chance of winning?

## 2. Get the olive out of the Martini glass (via Scientific American)

Rearrange these matchsticks so that the olive (that's the thing in the middle) is removed from the Martini glass. The olive must not be touched and you are only allowed to move two of the matchsticks. The Martini glass can be turned onto its left or right side, or even upside down, but must remain in the exact same shape.

None of the below images are the solution because the olive is still inside the glass or because the glass has taken a different form.

## 3. Draw one line on this equation to make it correct (via Sourish Jana)

## 4. 1,000 school lockers

There is a school with 1,000 students and 1,000 lockers. On the first day of term the headteacher asks the first student to go along and open every single locker, he asks the second to go to every second locker and close it, the third to go to every third locker and close it if it is open or open it if it is closed, the fourth to go to the fourth locker and so on. The process is completed with the thousandth student. How many lockers are open at the end?

## 5. Crazy cut (via Scientific American)

Add one cut (or draw one line), which doesn't need to be straight, that can divide this shape into two identical parts.

## 6. Colored socks puzzle (via The Guardian)

You are getting dressed in the dark and realise that you forgot to bind all your socks together into pairs. However you know there are exactly 10 pairs of white socks and 10 pairs of black socks in your draw. All the socks are exactly the same except for their color. How many socks do you need to take with you to ensure you have at least a pair that match?

## 7. Rays through the squares

Can you prove that angle C is the sum of angles A and B?

## 8. Love in Kleptopia (via Peter Winkler)

__The solutions __

## 1. How to beat Roger Federer at Wimbledon

It sounds obvious that you should ask to be ahead two sets to love (it takes 3 out of 5 sets to win

the men’s), and in the third set, ahead 5-0 in games and 40-love in the sixth game. (Probably you

want to be serving, but if your serve is like mine, you might prefer Roger to be serving the sixth

game down 0-40 so that you can pray for a double fault.)

Not so fast! These solutions give you essentially 3 chances to get lucky and win, but you can

get six chances—with three services by you and three by Roger. You still want to be up two sets

to none, but let the game score be 6-6 in the third set and 6-0—in your favor, of course—in the

tiebreaker.

Amit Chakrabarti of Dartmouth suggested the following improvement, based on the idea that

traditionally the complete score of a tennis match includes the game scores of all sets and, if the

game score was 6-6, the tiebreaker score as well. Then you could ask for example that the score be

6-0, 6-6 (9999-9997), 6-6 (6-0). The theory here (dubious, granted) is that while you were under

your magic spell, Roger was wearing himself out in the second-set tiebreaker and is now more likely

to blow one of the six upcoming match points.

## 2. Get the olive out of the Martini glass

Arrange the matchsticks like this:

## 3. Draw one line on this equation to make it correct

## 5+5+5+5=555

Solution 1:

Solution 2:

## 4. 1,000 school lockers

The only lockers that remain open are perfect squares (1, 4, 9, 16, etc) because they are the only numbers divisible by an odd number of whole numbers; every factor other than the number's square root is paired up with another. Thus, these lockers will be "changed" an odd number of times, which means they will be left open. All the other numbers are divisible by an even number of factors and will consequently end up closed.

So the number of open lockers is the number of perfect squares less than or equal to one thousand. These numbers are one squared, two squared, three squared, four squared, and so on, up to thirty one squared. (Thirty two squared is greater than one thousand, and therefore out of range.) So the answer is thirty one.

## 5. Crazy cut

The figure is cut into congruent halves like this:

## 6. Colored socks puzzle

Solution: Three socks.

With two socks it is possible to have one black and one white. 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.

## 7. Rays through the squares

## 8. Love in Kleptopia

This puzzle came from Caroline Calderbank, young daughter of mathematicians Ingrid Daubechies

and Rob Calderbank. In the solution she had in mind, Jan sends Maria a box with the ring in

it and one of his padlocks on it. Upon receipt Maria affixes her own padlock to box and mails it

back with both padlocks on it. When Jan gets it he removes his padlock and sends the box back to

Maria; voila! This solution is not just play; the idea is fundamental in Diffie-Hellman key exchange,

an historic breakthrough in cryptography.

Depending on one’s assumptions, other solutions are possible as well. My favorite was suggested

by several persons at the Gathering for Gardner VII: it requires that Jan find a padlock whose key

has a large hole, or at least a hole which can be sufficiently enlarged by drilling, so that the key can

be hooked onto a second padlock’s hasp.

Jan uses this second padlock, with the aforementioned key hooked on its hasp, to lock a small

empty box which he then sends to Maria. When enough time has passed for it to get there (perhaps

he awaits an email acknowledgment from Maria) he sends the ring in another box, locked by the first padlock. When Maria gets the ring box, she picks up the whole first box and uses the key

affixed to it to access her ring.

## *Could you work these out? How many did you get right?*

*Let us know in the comments below.*