## Numbers on a Dodecahedron

We have written the numbers from 1 to 12 on the faces of a regular dodecahedron. Then, we have written on each vertex the sum of the five numbers on the faces incident with it. Is it possible that 16 of these 20 sums are the same?

We color the vertices in five colors as shown in the image below, such that each face of the dodecahedron has 1 vertex of each color. Then, the sum of the four numbers from each color must be equal to the sum of all numbers written on the faces: 1 + 2 + … + 12 = 78.

If 16 of the vertices have the same number written on them, then by the pigeonhole principle there will be 4 vertices with identical colors and identical numbers. Since 78 is not divisible by 4, we conclude that this is impossible.

## Switch the Knights

Your goal is to switch the positions of the three white knights with the positions of the three black knights. What is the least number of moves required to do this?

The least number of moves required to switch the positions of the knights is 16. An example is shown below.

Next, we prove that it is impossible to switch the positions of the knights with less than 16 moves. Since the white knights occupy 2 black and 1 white squares, they need to end up on 2 white and 1 black squares, and each knight must make at least 2 moves in order to get to the opposite side, the total number of moves for the white knights should be an odd number, larger or equal to 2+2+2=6. The same applies to the total number of moves for the black knights. Therefore, the only possible way to get a total number of moves less than 16 is if both the white knights and the black knights move exactly 7 times.

We assume it is possible to switch the positions with 7+7=14 moves in total. Then, the white knight on A2 and one of the white knights on A1, A3 should make 2 moves each, and the third white knight should make 3 moves and land on a white square, either D1 or D3. Without loss of generality, we assume the knight on A1 makes 3 moves: A1-B3-C1-D3. Then, the knight on A2 should make 2 moves: A2-C3-D1, and the knight on A3 should make 2 moves: A3-B1-D2.

We make the same argument for the black knights. Since it is impossible that the white knight on A1 moves along the trajectory A1-B3-C1-D3 and also the black knight on D3 moves along the trajectory D3-C1-B3-A1, we conclude that the black knight on D1 should make 3 moves: D1-C3-B1-A3, the black knight on D2 should make 2 moves: D2-B3-A1, and the black knight on D3 should make 2 moves: D3-C1-A2.

This is only possible if:

1. the knight on A1 moves to C1 after the knight on D3 moves to A2
2. the knight on D3 moves to A2 after the knight on A2 moves to C3
3. the knight on A2 moves to C3 after the knight on D1 moves to B1
4. the knight on D1 moves to B1 after the knight on A3 moves to D2
5. the knight on A3 moves to D2 after the knight on D2 moves to B3
6. the knight on D2 moves to B3 after the knight on A1 moves to C1

We get a contradiction which means that the least number of moves is 16.

## Terrible Rebuses 2

Recognize the phrases depicted by the “terrible rebuses” below.

1. Blood is thicker than water.
2. Home is where the heart is.
3. It takes two to tango.
4. Haste makes waste.

## Clean Death

A man was going to bleach his socks because they had gotten muddy the day before. As he was pouring the bleach into the washing machine, he spilled some on the floor. He got some cleaning fluid and mopped it up with a rag. Minutes later he was dead. What killed him?

When the ammonia (NH3), found in cleaning fluids, is mixed with bleach (a dilute solution of sodium hypochlorite, NaOCl), a deadly gas (monochloramine, NH2Cl) is produced that can kill a person instantly. The resulting chemical reactions are the following:

1. Bleach decomposes, forming hydrochloric acid and subsequently chlorine gas:
NaOCl → NaOH + HOCl
HOCl → HCl + O
NaOCl + 2HClCl2 + NaCl + H2O
2. The chlorine gas reacts with the sodium hypochlorite in the bleach, releasing chloramine as a vapor:
2NH3 + Cl2 → 2NH2Cl

## In This Shoe…

Fill the three missing numbers (using words) in the shoe below.

You should fill the words FIVE, THREE, FOUR to get:

“In this shoe, there are five H’s, three O’s, and four T’s.”

## Bicycle Tracks

You are chasing a criminal riding a bicycle, and you find his tracks left in the dirt. By investigating the tracks, can you determine which direction the criminal has fled to: left or right?

The front wheels of the bicycle are traveling more distance and making sharper turns. Thus, the red tracks (as shown on the image below) correspond to the front wheel, and the green tracks correspond to the back wheel.

Then, since the back wheel is always pointing towards the front wheel, we conclude that the bicycle is moving towards the right.

## Terrible Rebuses 1

Recognize the phrases depicted by the “terrible rebuses” below.

1. No use over spilled milk.
2. Two wrongs don’t make a right.
3. The squeaky wheel gets the grease.
4. When there is a will, there’s a way.

## STANDARD Word

Given the word STANDARD, take away two letters and then add three digits to make a logical sequence.

Remove the two letters A to get ST-ND-RD and then add the digits 1, 2, 3 to get 1ST-2ND-3RD.

## 27 Figures of Speech

Find 27 figures of speech in the illustration by Ella Baron below.

1.    A piece of cake
2.    An ace up one’s sleeve
3.    Big cheese
4.    Play one’s cards close to one’s chest
5.    Cats got your tongue
6.    Cherry on the cake
7.    Crow over
8.    Cut the chase
9.    Don’t put all your eggs in one basket
10.    Hit the nail on the head
11.    Have (one’s) plate full
12.    Heart on one’s sleeve
13.    Holding a cat by the tail
14.    In a nutshell
15.    Kick the bucket
16.    Make ends meet
17.    Picture is worth a thousand words
18.    Joker in the pack
19.    Put a bug in someone’s ear
20.    Put one’s best foot forward
21.    Red herring
22.    Silver spoon in your mouth
23.    Spill the beans
24.    Tie in knots
25.    The shoe is on the other foot
26.    Time flies
27.    You can’t make an omelette without breaking egg

## Limbs

Partition the grid into disjoint “creatures”​, according to the following rules:

1. Each creature is defined as a shape of 4 connected branches that are each 1 cell wide.
2. For each creature, one of the branches ends up with a HEAD (always clued) and the other three branches end up with LIMBS (whenever clued, their directions matter).
3. A creature can never occupy a 2×2 region of cells and can never touch itself.

Examine the first example, then solve the other three puzzles.

The solutions are shown below.