## Five Points in a Square

There are 5 points in a square 1×1. Show that 2 of the points are within distance 0.75.

Split the unit square into 4 small squares with side lengths 0.5. At least one of these squares will contain 2 of the points. Since the diagonals of the small squares have lengths less than 0.75, these 2 points must be within such distance.

## Hidden Words Plants

Find all the hidden words in the grids.

The solutions are shown below.

## Fish in a Pond

There are 5 fish in a pond. What is the probability that you can split the pond into 2 halves using a diameter, so that all fish end up in one half?

Let us generalize the problem to N fish in a pond. We can assume that all fish are on the boundary of the pond, which is a circle, and we need to find the probability that all of them are contained within a semi-circle.

For every fish Fᵢ, consider the semi-circle Cᵢ whose left end-point is at Fᵢ. The probability that all fish belong to Cᵢ is equal to 1/2ᴺ⁻¹. Since it is impossible to have 2 fish Fᵢ and Fⱼ, such that the semi-sircles Cᵢ and Cⱼ contain all fish, we see that the probability that all fish belong to Cᵢ for some i is equal to N/2ᴺ⁻¹.

When N = 5, we get that the answer is 5/16.

## Thoka’s Rebus 4

Can you figure out which word is depicted by this rebus?

The first image on the first row depicts a TEAR. The second image depicts a RAT with a middle letter switched from A to I. From TEARRIT the A is removed, then PILLORY is added, and PILL is removed. The final answer is TERRITORY.

## Pronunciation Puzzles

The following 2 puzzles rely on misleading phrasing of the questions. Read them aloud to your friends and let them ponder upon them.

1. What has 4 letters, sometimes 9, and never 5
2. There are 30 cows and 28 chickens. How many didn’t?
3. Pronounce the following words: T-W-A, T-W-E, T-W-I, T-W-O
4. As I was walking across the London Bridge, I met a man.
He tipped his hat, and drew his cane.
In this riddle, I said his name. What is it?

The first puzzle is not a question. It is a statement, saying that the word “what” has 4 letters, the word “sometimes” has 9 letters, and the word “never” has 5 letters. There is nothing to solve, so the puzzle is figuring that out!

The second puzzle actually reads as “There are 30 cows and 20 ate chickens. How many didn’t?” Thus, the answer is that 10 cows didn’t eat chickens.

The third question often confuses people and they pronounce TWO as [twou] instead of [tuː].

The fourth riddle actually says: “He tipped his hat, ‘Andrew Hiscane'”. Thus, the name of the man is Andrew Hiscane.

## Rapunzel and the Prince

The evil witch has left Rapunzel and the prince in the center of a completely dark, large, square prison room. The room is guarded by four silent monsters in each of its corners.  Rapunzel and the prince need to reach the only escape door located in the center of one of the walls, without getting near the foul beasts. How can they do this, considering they can not see anything and do not know in which direction to go?

The prince must stay in the center of the room and hold Rapunzel’s hair, gradually releasing it. Then, Rapunzel must walk in circles around the prince, until she gets to the walls and finds the escape door.

## Remove TWO from FIVE

Wordplay: Remove TWO from FIVE and get FOUR.

Remove TWO letters, “F” and “E”, from the word “FIVE”, and get “IV”, which is FOUR in Roman numerals.

## White and Dirty

I am white when I am dirty, and black when I am clean. What am I?

## Is This Prime?

In the past few days, I, my friends, and a whole lot of Twitter people have been trying to beat each other’s scores in the game “Is This Prime?”.

The game itself is simple; you are shown random integers on the screen and you need to guess whether they are prime or composite. Since most presented numbers are between 1 and 200, after a couple of games, players naturally memorize them. However, this is a good opportunity for students to review some main number division rules.

1. Numbers that end with an even digit are divisible by 2. If the number formed by the last 2 digits of a number is divisible by 4, the original number is also divisible by 4. If the number formed by the last 3 digits of a number is divisible by 8, the original number is also divisible by 8.
• 536 is divisible by 2 because 6 is an even digit
• 1348 is divisible by 4 because 48 is divisible by 4
• 71824 is divisible by 8 because 824 is divisible by 8
2. Numbers that end with 5 are divisible by 5. Numbers that end with 25, 50, 75, or 00 are divisible by 25.
• 45 is divisible by 5
• 675 is divisible by 25
3. Numbers whose sum of digits is divisible by 3 are divisible by 3. Numbers whose sum of digits is divisible by 9 are divisible by 9.
• 144 is divisible by 3 because 1+4+4=9 is divisible by 3
• 1638 is divisible by 9 because 1+6+3+8=18 is divisible by 9
4. If the difference between the sum of the digits in odd places and the sum of the digits in even places is divisible by 11, the number is divisible by 11.
• 121 is divisible by 11 because 1+1-2=0 is divisible by 11
• 209 is divisible by 11 because 2+9-0=11 is divisible by 11
• 1628 is divisible by 11 because 1+2-6-8=-11 is divisible by 11
5. If the number before the last digit minus twice the last digit is divisible by 7, the original number is also divisible by 7.
• 161 is divisible by 7 because 16-2×1=14 is divisible by 7
• 371 is divisible by 7 because 37-2×1=35 is divisible by 7
• 1589 is divisible by 7 because 158-2×9=140 is divisible by 7

All the rules above apply in both directions, e.g. if the sum of the digits of a number is not divisible by 9, then the number itself is also not divisible by 9. There are more complicated rules that apply to larger numbers but the chances are you will never get to use them. If you are curious to learn more about them, go to the bottom of this article.

Once we know the main number division rules well, we are ready to play the game! Here are a few tips for getting high scores:

1. Memorize as many numbers as possible. Knowing the multiplication table up to 10×10, it should be easy to learn by heart whether each number up to 100 is prime or composite.
2. Pay attention to the last digit. If it is 5, then the number is composite (unless it is =5).
3. Check whether the sum of the digits is divisible by 3. If it is, then the number is composite (unless it is =3).
4. If the number is between 100 and 300, check whether the sum of the first and the third digits equals the second digit. If this is true, then the number is divisible by 11, and therefore it is composite. 209 is the only other number in this range divisible by 11.

Good luck playing and let us know if you beat our personal record of 67 points!

### Primes between 1 and 300:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199

211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293

### Sneaky composites:

51, 57, 87, 91

119, 133, 161, 169

209, 217, 221, 247, 253, 259, 287, 289, 299

## Some Math

Some of you may be wondering why the rules listed above work and whether we can create similar rules for larger numbers. Here are some explanations for the curious among you:

#### Rule for division by 3

Consider the 5-digit number ABCDE. It can be written as:

ABCDE = 10000A + 1000B + 100C + 10D + E

Since 10000 = 3 × 3333 + 1, 1000 = 3 × 333 + 1, 100 = 3 × 33 + 1, and 10 = 3 × 3 + 1, we can see that:

ABCDE = 3 × (3333A + 333B + 33C + 3D) + (A + B + C + D + E)

Therefore, ABCDE is divisible by 3 if and only if (A + B + C + D + E) is divisible by 3.

#### Rule for division by 11

Consider again the number ABCDE. Since 10000 = 11 × 909 + 1, 1000 = 11 × 91 – 1, 100 = 9 × 1 + 1, and 10 = 11 – 1, we can see that:

ABCDE = 11 × (909A + 91B + 9C + D) + (A – B + C – D + E)

Therefore, ABCDE is divisible by 11 if and only if (A – B + C – D + E) is divisible by 11.

#### Rule for division by 7

Once again, consider the number ABCDE. Notice that it can be written as:

ABCDE = 10 × ABCD + E

Now, let us find a number X such that 10X gives remainder 1 when divided by 7. Such number is X = 5. Indeed, 5 × 10 = 50 = 7 × 7 + 1. Therefore, the following statements are equivalent:

• ABCDE = 10 × ABCD + E is divisible by 7
• 5 × ABCDE = 49 × ABCD + ABCD + 5E is divisible by 7
• ABCD – 2E is divisible by 7

#### Rules for division by 13, 17, 19, etc.

The idea of the rule for division by 7 can be applied to rules for divisions by higher numbers. For example, here is how we can find a rule for division by 13:

1. Find the smallest positive integer X, so that 10X is divisible by 13. Such number is X = 4. Then, ABCDE is divisible by 13 if and only if 4 × ABCDE is divisible by 13.
2. Rewrite 4 × ABCDE as:
4 × ABCDE = 39 × ABCD + ABCD + 4E
3. Conclude that ABCDE is divisible by 13 if and only if ABCD + 4E is divisible by 13.

As an exercise, try to deduce a similar rule for division by 17!

For deeper understanding of how division of integers works, we recommend our more enthusiastic readers to look into Modular Arithmetic.

## Death in the Field

A man is lying dead in a field. Next to him, there is an unopened package. There are no other people or animals in the field. How did he die?

The man was sky diving and his parachute did not open.