Tear one off and scratch my head,
What once was red is black instead.

What is it?

## Does Not Divide Another

How many numbers between 1 and 100 can you pick at most, so that none of them divides another?

You can choose at most 50 numbers: 51, 52, 53, …, 100.

In order to see that you cannot choose more than 50, express each number in the form 2ⁿ×m, where m is an odd number. Since no 2 numbers can have the same m in their expressions, and there are only 50 odd numbers between 1 and 100, the statement of the problem follows.

## Big on Saturday

I am big on Saturday and Sunday.
I am small on Tuesday, Wednesday, and Thursday.
I am not on Monday and Friday.
What am I?

The answer is the letter S.

## Surrounded by Rooks

White to play and mate Black in 20 moves.

The moves are as follows:

1. Qb7+ Rc5c6
2. Qa5+ Rc4c5
3. Qb3+ Rd4c4
4. Qd2+ Re4d4
5. Qf3+ Re5e4
6. Qg5+ e5
7. Qf7+ Rd6e6
8. Qd8+ Rc6d6
9. Qb7+ Rc5c6
10. Qa5+ Rc4c5
11. Qb3+ Rd4c4
12. Qd2+ Re4d4
13. Qf3+ e4
14. Qg5+ Re6e5
15. Qf7+ Rd6e6
16. Qd8+ Rc6d6
17. Qb7+ Rc5c6
18. Qa5+ Rc4c5
19. Qb3+ Rd4c4
20. Qd2x

## Murder Mystery Puzzles

In the three murder cases below, you can read the testimonies of all suspects. For each case, find who the killer is, knowing that no 2 people are in the same row or column, and that the killer was alone in a room with the victim.

The solutions are shown below.

## Murdered Wife

One night, a man received a call from the Police. The Police told the man that his wife was murdered, and that he should get to the crime scene as soon as possible. The man immediately hung up the phone and drove his car for 20 minutes. As soon as he got to the crime scene, the Police arrested him, and he got convicted for murder.

How did the Police know that the man committed the crime?

The police did not tell the man where the crime scene was.

## What Does the Liаr Do

What does the liar do when he dies?

He lies still.

## Self-Describing Number

Find all 10-digit numbers with the following property:

• the first digit shows the number of 0s in the number
• the second digit shows the number of 1s in the number
• the third digit shows the number of 2s in the number, and so on

Let the number be ABCDEFGHIJ. The number of all digits is 10:

A + B + C + D + E + F + G + H + I + J = 10

Therefore, the sum of all digits is 10. Then:

0×A + 1×B + 2×C + 3×D + 4×E + 5×F + 6×G + 7×H + 8×I + 9×J = 10

We see that F, G, H, I, J < 2. If H = 1, I = 1, or J = 1, then the number contains at least 7 identical digits, clearly 0s. We find A > 6 and E = F = G = 0. It is easy to see that this does not lead to solutions, and then H = I = J = 0.

If G = 1, we get E = F = 0. There is a 6 in the number, so it must be A. We get 6BCD001000, and easily find the solution 6210001000.

If G = 0, F can be 0 or 1. If F = 1, then there must be a 5 in the number, so it must be A. We get 5BCDE10000. We don’t find any solutions.

If F = 0, then the number has at least five 0s, and therefore A > 4. However, since F = G = H = I = J = 0, the number does not have any digits larger than 4, and we get a contradiction.

Thus, the only solution is 6210001000.

## 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.

## 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.