Raymond Merrill Smullyan (1919 – 2017) was an American mathematician, magician, concert pianist, logician, Taoist, and philosopher. Born in Far Rockaway, New York, his first career was stage magic. He earned a BSc from the University of Chicago in 1955 and his Ph.D. from Princeton University in 1959. He is one of many logicians to have studied with Alonzo Church.
If you make a CORRECT statement, you will get either a lollipop or a chewing gum. If you make a FALSE statement, you will get either a chocolate or a car. What statement should you make in order to get the car?
You should say “I will receive a chocolate”. This statement cannot be correct, since if it was, you would get a lollipop or a chewing gum, not a chocolate. Therefore, you will get the car.
Below each of the following Venn diagrams there are seven tiles consisting of two letters. Place each tile in a different region so that the four tiles in each circle can be rearranged to solve the corresponding clue.
The solutions are shown below.
Add the 25 letters between A and Y to the grid below. Each one should appear exactly once. When you have finished, you must be able to spell the following words moving horizontally, vertically or diagonally around the grid:
AXIOM, BANQUET, CHOMP, FAWN, JOY, LURKED, RUN, STY, VYING
The solution is shown below:
How many matchsticks do you need to remove so that no squares of any size remain?
Nine matchsticks are enough, as seen from the solution below.
To see that eight matchsticks are not enough, notice that removing an inner matchstick reduces the number of 1×1 squares at most by 2. Since there are 16 such small squares, in order to get rid of them all, we need to remove only inner matchsticks. However, in this case, the large 4×4 square will remain.
Place arrows along the hexagon edges so that the number of arrows pointing to each hexagon equals the number of dots inside, adhering to the following rules:
The solutions are shown below.
Place the numbers from 1 to 8 on the vertices of a cube so that the sum of the four numbers on every face is the same.
The solution is shown below:
How many pawns can you place at most on a chess board so that no three pawns lie on a single line, horizontal, vertical, or diagonal?
Since there are 8 rows on the chess board, if you place more than 16 pawns, then there will be at least 3 that lie on a single horizontal line. An example with 16 pawns is shown below:
How many X’s can you place the most on a tic-tac-toe board without getting 3 X’s on a line?
You can place 6 X’s without getting 3 X’s on a line, as shown below.
A man describes his daughters, saying, “They are all blonde but two; all brunette but two; and all redheaded but two.” How many daughters does the man have?
The man has 3 daughters; one blonde, one brunette, and one redhead.
You have 12 balls, 11 of which have the same weight. The remaining one is defective and either heavier or lighter than the rest. You can use a balance scale to compare weights in order to find which is the defective ball and whether it is heavier or lighter. How many measurement do you need so that will be surely able to do it?
It is easy to see that if we have more than 9 balls, we need at least 3 measurements. We will prove that 3 measurements are enough for 12 balls.
We place 4 balls on each side of the scale. Let balls 1, 2, 3, 4 be on the right side, and balls 5, 6, 7, 8 on the left side.
CASE 1. The scale does not tip to any side. For the second measurement we place on the left side balls 1, 2, 3, 9 and on the right side balls 4, 5, 10, 11.
If the scale again does not tip to any side, then the defective ball is number 12 and we can check whether it is heavier or lighter with our last measurement.
If the scale tips to the left side, then either the defective ball is number 9 and is heavier, or it is number 10/11 and is lighter. We measure up balls 10 and 11 against each other and if one of them is lighter than the other, then it is the defective one. If they have the same weight, then ball 9 is the defective one.
If the scale tips to the right side, the procedure is similar.
CASE 2. Let the scale tip to the left side during the first measurement. This means that either one of the balls 1, 2, 3, 4 is defective and it is heavier, or one of the balls 5, 6, 7, 8 is defective and it is lighter. Clearly, balls 9, 10, 11, 12 are all genuine. Next we place balls 1, 2, 5, 6 on one side and balls 3, 7, 9, 10 on the other side.
If the scale tips to the left, then either one of the balls 1, 2 is defective and it is heavier, or ball 8 is defective and lighter. We just measure up balls 1 and 2 against each other and find out which among the three is the defective one.
If the scale tips to the right, the procedure is similar.
If the scale does not tip to any side, then either the defective ball is 4 and it is heavier, or the defective ball is 8 and it is lighter. We just measure up balls 1 and 4 against each other and easily find the defective ball.
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