If the first image below is a side view, and the second image is the front view (as viewed from the arrow). What’s the top view of this object?

[**Source:** Rob Eastaway]

A compendium of idle musings

If the first image below is a side view, and the second image is the front view (as viewed from the arrow). What’s the top view of this object?

[**Source:** Rob Eastaway]

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Posted in Puzzles

There are 12 coins, one of which is a counterfeit. The false coin differs in weight from the true ones, but you don’t know whether it’s heavier or lighter. Find the counterfeit coin using three weighings in a pan balance.

[From ** Futility Closet**]

Posted in Puzzles

From here.

Posted in Puzzles, Recreational Math

[Source: *DataGenetics*, Pic Courtesy: CREATTICA]

Posted in Puzzles

Shurl and Watts, at a base on Pluto, are in charge of distributing doyles to more distance outposts. Doyels are the size of peas, all identical, each weighing precisely 1 gram. They are indispensable in hyperspace propulsion systems.

Doyles come in cans of 100 doyles each, and shipments are made up of six cans at a time. The Pluto base has a sensitive spring scale capable of registering fractions of milligrams.

One day, a week after a shipment of doyles, a radio message came from the manufacturing company in Hong Kong, “Urgent. One can is filled with defective doyles, each with an excess weight of 1 milligram. Identify the can and destroy its doyles at once.”

“I suppose,” said Watts, “we’ll have to make six weighings, one doyle from each can.”

“Not so, my dear Watts,” said Shurl. “we can identify the can of defectives with just one weighing.” [And then he goes on to explain how this can be achieved using a single weighing.]

“How absurdly simple!” exclaimed Watts, while Shurl shrugged.

A month later, after the next shipment, another message arrived: “Any of the six cans, perhaps all of them, may be full of defective doyles, each 1 milligram overweight. Identify and destroy all defective doyles.”

“This time,” said Watts, “I suppose we’ll have to weight separately a doyle from each can.”

Shurl put his fingertips together and gazed at a picture of Isaac Asimov on the wall. “A capital problem, Watts. No, I think we can still do it in just one weighing.”

What algorithm does Shurl have in mind?

[From *Mathematical Puzzle Tales* by Martin Gardner]

**PS:** The solution of the first weighing problem (one defective can) is actually provided in the original text of this puzzle. I’ve removed it for those who are not familiar with the solution and want to give it a shot.

Posted in Puzzles

Here’s a puzzle:

You find yourself on an unknown uninhabited planet. You look up the sky and notice a giant star that looks like our sun. It rises in one particular direction and sets in the opposite, so you figure that this planet (that you’re on) orbits around this particular star. Now it’s easy to define a day on this planet: it’s the time between two subsequent sunrises. The question is: how would you define a year on this planet? (Assuming, of course, that your survival on this planet is not an issue.)

Someone posted this question on Quora, which got me thinking about the definition of a year. One earth-year is defined as 365 days, or 12 moon cycles, and these measures are not arbitrary. They (approximately) represent the time it takes for the earth to complete one orbit around the sun. With that in mind, the answer to the above puzzle is rather straight-forward for a planet that has an **axial tilt**.

If you landed on a planet with an axial tilt then the sun’s trajectory in the sky will change over time and you can track it to measure one year. The following approach can be used to measure one complete orbit: Mark the point at the horizon where the sun rises the next morning. Every subsequent morning the sun will rise at a slightly different point. (The amount of shift depends on the degree of axial tilt and how fast the planet is orbiting its sun.) For a while, the sunrise points will shift away from your initial mark in one particular direction (either left or right). And after a certain period, like a pendulum, they will start moving back towards the initial mark. (This U-turn marks a solstice on this planet.) It will keep moving and bypass the initial mark, move in the opposite direction, and then take another U-turn to come back to the initial mark. The morning when sun rises again from the original reference point would mark the completion of one year on this planet. (Note that this description changes slightly if you started observing the sunrises exactly on the day of one of the two solstices on this planet.)

Here’s how the sun’s trajectory looks like from Earth (looking southward from the Northern hemisphere):

In our solar system, most planets have an axial tilt. Earth’s current axial tilt of 23.4° is responsible for seasons, rain, and consequently, the existence of life on Earth. The most curious among these are Venus and Uranus; they are the only planets that rotate clockwise (while looking at the solar system from the top). Venus has flipped almost completely and is upside down as compared to other planets. While Uranus, with an obliquity of 98°, would look like a tilted rolling ball — as opposed to all other planets that look like tilted spinning tops.

Back to the question of measuring one year, if you landed on a planet *without* an axial tilt, I have no idea how you would measure a year.

Suppose I offer you two prizes: prize * A* and prize

If the statement is clearly true, I will give you a prize. You will win either * A* or

And if your statement is false, you won’t get any prize.

It’s clear from this that you want to make a statement that is clearly true, because otherwise you won’t win any prize. You can say something like “One plus one equals two.” This is clearly true, and you will win either * A* or

However, let’s say you really would like to win prize * A*. The question is: what statement can you make that will ensure that you will win prize

Posted in Puzzles

Over the years, I have expanded my Rubik’s-cube-inspired-puzzle collection by adding several cool variants of the original cube that was made by the Hungarian architect in the late 70’s. My collection of these twisty puzzles now boast a wide variety:

- The classic Rubik’s cube,
- 30th anniversary limited edition wood cube,
- LanLan Diamond (Octahedron) puzzle,
- 4×4 Mirror cube,
- Dayan V2 color cube,
- Meffert’s Gear cube,
- 2×3 puzzle,
- 2×2 cube,
- 1×3 (Floppy) puzzle,
- 3×4 puzzle,
- Rubik’s Cheese,
- Mozhi Tetris block,
- 3×3 Mirror cube,
- QJ Pyraminx,
- another Rubik’s cube (with a silver side instead of white),
- 4×4 cube,
- another 2×3 puzzle, and
- Void cube.

The oldest one among this lot is #15, you can see some stickers starting to wear off. I have yet to solve a couple of these, but my favorite so far is the 3×3 Mirror cube (#13 above). The irregularities in the shape of its 26 “cubies” keeps me busy for at least 10 to 15 minutes every time I play with it. In addition to the irregular shapes, each side has the same color which adds to the complexity and makes it more difficult to solve than a Rubik’s cube.

Five pirates find 100 gold coins in their recent loot. In order to distribute the loot among them, they use the following approach:

The most senior pirate will propose a distribution. If at least half of them agree with the proposed distribution, the loot will be divided as proposed. Otherwise the most senior pirate (who made the proposal) will be killed. Then they start over with the next most senior pirate, and so on.

The question is: what distribution should the most senior pirate propose so that he won’t be killed? Assume that the pirates are intelligent, greedy and prefer not to die.

**Update:** The solution is posted in the comments section.

Posted in Puzzles

There are several different versions of this particular paradox – like *Holli’s Paradox*, and *The Unexpected Hanging Paradox*. Before I present the surprise examination version, it behooves me to mention that no correct (or final) solution to this paradox has been established yet.

Here’s the paradox:

A teacher announces in class that an examination will be held on some day during the following week, and moreover that the examination will be a surprise.

The students argue that a surprise exam cannot occur. For suppose the exam were on the last day of the week. Then on the previous night, the students would be able to predict that the exam would occur on the following day, and the exam would not be a surprise. So it is impossible for a surprise exam to occur on the last day. But then a surprise exam cannot occur on the penultimate day, either, for in that case the students, knowing that the last day is an impossible day for a surprise exam, would be able to predict on the night before the exam that the exam would occur on the following day. Similarly, the students argue that a surprise exam cannot occur on any other day of the week either.

Confident in this conclusion, they are of course totally surprised when the exam occurs (on Wednesday, say). The announcement is vindicated after all. Where did the students’ reasoning go wrong?

***

Similar posts: *The Monty Hall Paradox, A Mathematical Conundrum, The Voting Paradox, Broken Clocks.*

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