## 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.