Oct 19

## Two Lost Cards in a Deck

Below you can read the steps of a magic trick, as well as a video of its live performance. Your goal is to figure out how the trick is done, then perform it for your friends and challenge them to figure out the trick themselves.

1. Take out from your pocket a deck of cards, which is visibly shuffled.
2. Ask your first assistant to cut the deck, then take the top card from the bottom pile of cards and memorize it.
3. Ask your second assistant to take the next card from the bottom pile and memorize it.
4. Ask your first assistant to return his card back on the top of the bottom pile, then ask your second assistant to do the same.
5. Place the two piles of cards on top of each other and cut the deck multiple times.
6. Split the deck into two piles of cards, dealing consecutively one card on the left, then one card on the right, and so on, until you run out of cards.
7. Take one of the two piles of cards, look at it, and guess correctly what cards were chosen by your assistants.

How does the magic trick work? Below you can see a live performance of the magic trick from Penn and Teller’s show Fool Us.

The secret of the trick is to memorize the group of cards which are located in even positions and the group of cards which are located in odd positions in the original deck. An easy way for doing this is to split the cards into two groups, such that the cards in the first group are only spades and diamonds, and the cards in the second group are only clubs and hearts.

When the two assistants pick their cards and then return them back into the deck, the order of the cards is reversed. When you split the original deck into two piles (even after cutting it several times), each of the piles will contain a card which should not be there. For example, the group of spaes and diamonds will contain one clubs card, and the group of clubs and hearts will contain one diamonds card. These two cards are the ones which were picked by the assistants.

Oct 17

## Ten Lanterns

You have ten lanterns, five of which are working, and five of which are broken. You are allowed to choose any two lanterns and make a test which tells you whether there is a broken lantern among them or not. How many tests do you need until you find a working lantern?

Remark: If the test detects that there are broken lanterns, it does not tell you which ones and how many (one or two) they are.

You need 6 tests:

(1, 2) → (3, 4) → (5, 6) → (7, 8) → (7, 9) → (8, 9)

If at least one of these tests is positive, then you have found two working lanterns.

It all of these tests are negative, then lantern #10 must be working. Indeed, since at least one lantern in each of the pairs (1, 2), (3, 4), (5, 6) is not working. Therefore, there are at least 2 working lanterns among #7, #8, #9, #10. If #10 is not working, then at least one of the pairs (7, 8), (7, 9), or (8, 9) must yield a positive test, which is a contradiction.

With some case analysis, it is not hard to see that 5 tests are not enough.

Oct 11

## Diagonal in a Rectangle

A 1000 × 1004 rectangle is split into 1 × 1 squares. How many of these squares does the main diagonal of the large rectangle pass through?

Notice that the number of small squares the main diagonal passes through is equal to the number of horizontal and vertical lines it intersects. Indeed, every time the diagonal goes through the interior of one square to the interior of another, it must intersect one of these lines.

There are 1000 + 1004 = 2004 lines which are intersected by the main diagonal. However, on four occasions (which is the greatest common divisor of 1000 and 1004), the main diagonal intersects one horizontal and one vertical line at the same time, which results in double-counting., so we must subtract 4 from the answer.

Therefore, the answer is 1000 + 1004 – 4 = 2000.

Oct 3

## The Next Die

Which die completes the sequence?

If you look at dots in the top row, you will see that they are put together into groups of 1, 2, 3, 4, and so on.

Sep 25

## Rhombuses

A regular hexagon is split into small equilateral triangles and then the triangles are paired arbitrarily into rhombuses. Show that this results into three types of rhombuses based on orientation, with equal number of rhombuses from each type.

Color the rhombuses based on their type and imagine the diagram represents a structure of small cubes arranged in a larger cube. If you look at the large cube from three different angles, you will see exactly the three types of rhombuses on the diagram.

Alternatively, the problem can be proven more rigorously by considering the three sets of non-intersecting broken lines connecting the pairs of opposite sides of the hexagon, as shown on the image below. The type of each rhombus is determined by the types of the broken lines passing through it. Therefore, there are n² rhombuses of each type, where n is the length of the hexagon’s sides.

Sep 21

## 0 > 2, 2 > 5, 5 > 0

0 > 2, 2 > 5, 5 > 0. What is this?

This is the game “Rock, Paper, Scissors”. Rock (0 fingers) beats scissors (2 fingers). Scissors (2 fingers) beats paper (5 fingers). Paper (5 fingers) beats rock (0 fingers).

Sep 19

## Four Points in the Plane

Find all configurations of four points in the plane, such that the pairwise distances between the points take at most two different values.

Coming soon.