Tag: Advanced

Ambiguous Clock

Category: Insight
Difficulty:

The hands of my alarm clock are indistinguishable. How many times throughout the day their positioning is such that one cannot figure out which is the hour hand and which is the minute hand?

Remark: AM-PM is not important.

Imagine that you have a third hand which moves 12 times as fast as the minute hand. Then, at any time, if the hour hand moves to the location of the minute hand, the minute hand will move to the location of the imaginary hand. Therefore, our task is to find the number of times during the day when the hour hand and the imaginary hand are on top of each other, and the minute hand is not.

Since the imaginary hand moves 144 times faster than the hour hand, the two hands are on top of each other exactly 143 times between 12AM and 12PM. Out of these 143 times, 11 times all three arrows are on top of each other. Therefore, we have 2 × (143 – 11) = 264 times when we cannot figure out the exact time during the entire 24-hour cycle.

Unknown Author January 28, 2020
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Ten Lanterns

Category: Deduction
Difficulty:

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.

Unknown Author December 28, 2019
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Diagonal in a Rectangle

Category: Mathematics
Difficulty:

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.

Unknown Author December 19, 2019
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Rhombuses

Category: Insight
Difficulty:

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.

Unknown Author December 13, 2019
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