Category Archives: Brain Teasers


A man was moving to a new house. He rented a moving truck, put all his belongings in it, and drove to his new place. He entered the garage with the truck and took all his belongings out of the truck. When he tried to exit the garage with the truck, he couldn’t. Why?

The empty truck was just slightly taller than the garage door. When it was packed with items, the truck’s height got lower, so the man could enter the garage. Once the items were unpacked, the truck was once again taller than door, so it couldn’t get out.


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.


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.


The sentence below is grammatically correct. Can you explain it?

Buffalo buffalo Buffalo buffalo buffalo buffalo Buffalo buffalo.

The sentence says that buffalo (animals) from Buffalo (city, US), which are buffaloed (intimidated) by Buffalo (city, US) buffalo (animals), themselves buffalo (intimidate) buffalo (animals) from Buffalo (city, US).


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.


The sides of a rectangle have lengths which are odd numbers. The rectangle is split into smaller rectangles with sides which have integer lengths. Show that there is a small rectangle, such that all distances between its sides and the sides of the large rectangle have the same parity, i.e. they are all even or they are all odd.

Source: Shortlist IMO 2017

Split the large rectangle into small 1×1 squares and color it in black and white, chessboard-style, such that the four corner squares are black. Since the large rectangle has more black squares than white squares, one of the smaller rectangles also must have more black squares than white squares. Therefore, the four corners of that smaller rectangle are all black. Then, it is easy to see that all distances between its sides and the sides of the large rectangle have the same parity.