Author: Unknown Author

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At Creepy Beasts Inc., three of the most dreaded animals, a tiger, a wolf, and a bear, sat in their boardroom in silence while they awaited their boss. Then, Mr. Tiger broke the silence.

“Isn’t it odd that our surnames match our species, yet none of our surnames match our own?”

The wolf replied, “Yeah, but does anyone care?”

They sat in silence again…

Can you figure out the surname of each animal?

Since the wolf replied to Mr. Tiger, his surname can be neither Tiger nor Wolf. Therefore, the wolf’s surname is Mr. Bear. Subsequently, Mr. Tiger must be a bear, and finally, Mr. Wolf must be a tiger.

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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.

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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.

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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).

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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.

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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.