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There are two cogwheels on a table. The bigger one has 10 teeth and is fixed to the table. The smaller one has 5 teeth and revolves around the bigger one. If the smaller cogwheel makes one full rotation around the bigger cogwheel, how many rotations will it make with respect to the table?
The answer is three rotations in total. Two because of the ratio 10:5, one more because of the movement of the smaller cogwheel.
Get from START to END in this planet maze, created by Sean Jackson.
The solution is shown below.
What reptile does this sum spell?
What mammal does this sum spell?
TREE + LOAD – REEL = TOAD
BOA + BEER – BEE = BOAR
White plays and mates Black in one move. However, there is a mystery in this position that has to be revealed first.
The mystery is that someone has just placed one extra black pawn on the board – there are 9 in total. Also, no matter which one is the added pawn, there always exists a mate in one move.
If the extra pawn was a7 – Qb6
If the extra pawn was b7 – Kc6
If the extra pawn was c4 – Qb4
If the extra pawn was d3 – Qe4
If the extra pawn was e3 – Bxf2
If the extra pawn was f7 – Ke6
If the extra pawn was g6 – Rg4
If the extra pawn was h3 – Rh4
A perfectly symmetrical square 4-legged table is standing in a room with a continuous but uneven floor. Is it always possible to position the table in such a way that it doesn’t wobble, i.e. all four legs are touching the floor?
The answer is yes. Let the feet of the table clockwise are labeled with 1, 2, 3, 4 clockwise. Place the table in the room such that 3 of its feet – say 1, 2, 3, touch the ground. If foot 4 is on the ground, then the problem is solved. Otherwise, it is easy to see that we can not put it there if we keep legs 2 and 3 in the same places. Now start rotating the table clockwise, keeping feet 1, 2 and 3 on the ground at all times. If at some point foot 4 touches the ground as well, the problem is solved. Otherwise, continue rotating until foot 1 goes to the place where foot 2 was and foot 2 goes to the place where foot 3 was. Foot 3 will be on the ground, but this contradicts the observation that initially we couldn’t place legs 2, 3 and 4 on the ground without replacing feet 2 and 3.
Three professors fell asleep under a tree. At some point a prankster passed by and painted their faces with black dye. When the professors woke up, each of them saw the others’ faces and started laughing at them. After a while though, they stopped laughing, realizing that their own faces were painted as well. How did they deduce that?
Let us denote the professors with A, B and C. The thought process of A would go like this: “If my face is not painted, then B will see that C is laughing at him and will realize immediately that he is being pranked. However, B was laughing for a while and therefore my I must being pranked as well.”. Then A will stop laughing and the same will happen with the other two professors B and C.
You have two groups of words:
To which group does “repetitive” belong?
The first group contains self-explanatory words (known as autologicals), the second group does not. Therefore “repetitive” should belong to the first group.
When young, I am sweet in the sun.
When middle-aged, I make you gay.
When old, I am valued more than ever.
What am I?
The answer is WINE.
Louis has a bar of chocolate 4×6 which is marked into 24 little squares. At each step, he breaks up one of its pieces along any of the marked horizontal/vertical lines. Show that no matter how he does that, it will always take the same number of steps until the chocolate is broken into single 1×1 pieces.
Every time he splits the chocolate, the number of pieces increases by 1. Therefore it will always take him 23 steps to split it into single pieces.
Get from START to END in this maze, created by Sean Jackson.
The solution is shown below.
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