Circular Racetrack

Suppose you are on a one-way circular racetrack. There are 100 gas cans randomly placed on different locations of the track and the total amount of gas in the cans is enough for your car to complete an entire circle. Assume that your car has no gas initially, but you can place it at any location on the track, then pick up the gas cans along the way and use them to fill the tank. Show that you can always choose a starting position so that you can complete an entire circle.

Imagine you put your car at any location, but instead with an empty tank, you start with enough gas to complete the circle. Then, simply track the amount of gas you have and locate the point on the track where it is the lowest. If you choose that location as a starting point, you will be able to complete the track.

Avoiding Bad Luck

You are walking alone on the sidewalk. There are no stars on the sky, no moonlight, all of the lamps on the street are broken, you don’t carry any source of light with you and there aren’t any cars or other people approaching. A silent black cat tries to cross your way, but you somehow spot it and turn around in order to avoid bad luck. How did you see the cat?

All of this happened during a (cloudy) day.

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Sum Up to 15

Tango and Cash are playing the following game: Each of them chooses a number between 1 and 9 without replacement. The first one to get 3 numbers that sum up to 15 wins. Does any of them have a winning strategy?

Place the numbers from 1 to 9 in a 3×3 grid so that they form a magic square. Now the game comes down to a standard TIC-TAC-TOE, and it is well-known that it always leads to a draw when both players use optimal strategies.

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David Copperfield

David Copperfield and his assistant perform the following magic trick. The assistant offers a person from the audience to pick 5 arbitrary cards from a regular deck and then hands them back to him. After the assistant sees the cards, he returns one of them to the audience member and gives the rest one by one to David Copperfield. After the magician receives the fourth card, he correctly guesses what card the audience member holds in his hand. How did they perform the trick?

Out of the five cards, there will be (at least) two of the same suit; assume they are clubs. Now imagine all clubs are arranged in a circle in a cyclic manner – A, 2, 3, … J, Q, K (clock-wise), and locate the two chosen ones on it. There are two arks on the circle which are connecting them and exactly one of them will contain X cards, with X between 0 and 5. Now the assistant will pass to David Copperfield first the clubs card which is located on the left end of this ark, will return to the audience member the clubs card which is located on the right end of it and, with the remaining three cards, will encode the number X. In order to do this, he will arrange the three extra cards in increasing order – first clubs A-K, then diamonds A-K, then hearts A-K and finally spades A-K. Let us call the smallest card in this order “1”, the middle one “2” and the largest one “3”. Now, depending on the value of X, the assistant will pass the cards “1”, “2” and “3” in the following order:

X=0 ⇾ 1, 2, 3
X=1 ⇾ 1, 3, 2
X=2 ⇾ 2, 1, 3
X=3 ⇾ 2, 3, 1
X=4 ⇾ 3, 1, 2
X=5 ⇾ 3, 2, 1

In this way David Copperfield will know the suit of the audience member’s card and also with what number he should increase the card he received first in order to get value as well. Therefore, he will be able to guess correctly.

12 Balls, 1 Defective

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.

Not Twins

A teacher enters the classroom and sees on the first row two students sitting next to each other, looking completely identical. She asks them if they are twins, but the students simultaneously reply that they are not. After checking in the records, the teacher furthermore discovers that the two children have the same mother and father. What is the explanation?

The students are three of a triplet.