From Henry Dudeney:

Imagine a man going to the North Pole. The points of the compass are, as everyone knows:

He reaches the Pole and, having passed over it, must turn about to look North. East is now on his left-hand side, West on his right-hand side, and the points of the compass therefore

… which is absurd. What is the explanation?

We’ve seen chess problems in which White must mate in half a move and even in -1 moves.

In this one White must mate in 0 — he must deliver checkmate without touching any of his pieces:

How can he do this? Imagine that the position arose in an actual game.

A bit of conjuring adapted from Augustus de Morgan:

1. Think of a one-digit number and remember it. (Example: 4.)

2. Write down a number of any length. Jumble the figures into another number, and subtract one from the other:

3. Count the letters in your father’s first name, your state capital, and the name of your favorite Beatle, and add them together.

4. Multiply this number by 4 and its reverse by 5. Add these together, plus the number from step 1.

For example, suppose your father’s name is William, your state capital is Oklahoma City, and you choose Paul. That’s 23 letters in all, and 23 reversed is 32. (4 × 23) + (5 × 32) + 4 = 256.

5. Mix these figures (256) into the result from step 2 (4600708659), in any order, say 4560207086569.

Seeing nothing but this final list of figures, the conjurer names the one-digit number from step 1.

How does he do it?

From Henry Dudeney:

Here is a little military puzzle that may not give you a moment’s difficulty. It is such a simple question that a child can understand it and no knowledge of artillery is required. Yet some of my readers may find themselves perplexed for quite five minutes.

An inventor offered a new large gun to the committee appointed by our government for the consideration of such things. He declared that when once loaded it would fire 60 shots at the rate of a shot a minute. The War Office put it to the test and found that it fired 60 shots an hour, but declined it “as it did not fulfill the promised condition.”

“Absurd,” said the inventor, “for you have shown that it clearly does all that we undertook it should do.”

“Nothing of the sort,” said the experts. “It has failed.”

Can you explain this extraordinary mystery? Was the inventor, or were the experts, right?

Point A is the circle’s center. What is its radius?

Amy, Bob, Cindy, and Dave are the last four colonists on Mars, which is being overrun by Hideous Sand Beetles. The last evacuation ship leaves in 16 minutes. To reach it, they must pass through the Soul-Freezing Tunnel of Yx, which can accommodate only two people at a time. And anyone passing through the tunnel must light his way with the Fabulous Oracle of Zeb. (Mars is very dramatic.)

• Amy can travel the tunnel in 1 minute.
• Bob can travel the tunnel in 2 minutes.
• Cindy can travel the tunnel in 5 minutes.
• Dave can travel the tunnel in 8 minutes.

This is a problem. If Amy and Dave go first, they’ll reach the other side in 8 minutes (Dave’s top speed). If Amy then runs back and escorts Cindy, and then Bob, she and Bob will be only halfway through the tunnel when the ship departs. Are they doomed?

In 1905, the Ingersoll company engaged Sam Loyd to invent a puzzle to promote its new dollar watch. Loyd sent the illustration above and asked:

How soon will the hour, minute and second hands again appear equal distances apart?

The company advertised the puzzle in Scribner’s in June, promising a free watch to the first 10,000 correct respondents. “The full problem is stated above,” the ad ran, “and no further information can be given in fairness to all contestants.” Further, it said, Loyd’s solution “is locked in our safe, inaccessible to any one.”

Perhaps it still is — I can’t find any record of a solution to the puzzle. I offer it here for what it’s worth.

06/02/2024 UPDATE: Reader Alexander Rodgers writes:

I had to give it a go. To summarise my findings:

• It isn’t actually possible for the three hands of a clock to be perfectly equally spaced apart, barring some unusual hand-moving mechanisms. Even the time shown in the picture isn’t perfectly equal.
• The answer therefore probably hinges on the phrase “appear equal”.
• That’s subjective, so I present three answers, and one can pick which is the earliest time that “appears” equal:
• 0h 43m 23s later, at 3:37:58
• 2h 54m 34s later, at 5:49:09
• 6h 10m 50s later, at 9:05:25.

The company may claim the puzzle is present in full. But on attempting it, I found it difficult to determine what assumptions the puzzle is working under.

I started by assuming two things:

1. That “equal distances apart” means each pair of hands forms an angle of exactly 120 degrees.
2. That the hands move continuously, rather than ‘ticking’ in some way.

Under these assumptions though, the example time pictured isn’t actually at equal distances.

The time pictured is about 2:55:35. I found the exact time at which the second and minute hands are exactly 120º: it is 2:55.34.576… (Specifically, it is (171+2/3)/59 hours after twelve.) At this time:

The hour hand has moved (2 + 54/60 + 34.576/3600)/12 revolutions or 87.29º;

The minute hand has moved 2 full revolutions, plus another (54/60 + 34.576/3600) or 327.46º;

The second hand has moved many revolutions plus 34.576/60 or 207.46º.

So the three angles are 120º exactly, 120.17º, and 119.83º. Any times close to this time might get another pair bang on but the second/minute pair will be off.

In fact, I have determined that under my two assumptions there are no points in time at all where the hands truly are equal distances apart.

Proof:

Measure time (t) in “hours past twelve”. Then the number of revolutions of each hand at time t is:

S(t) = 60t for the second hand,

M(t) = t for the minute hand, and

H(t) = (1/12)t for the hour hand.

We can take the difference in any two of these functions. If the result is a whole number plus 1/3, then the angle between them is 1/3 of a full circle or 120º. If the result is a whole number plus 2/3, then the angle between them is also 120º. But if it’s anything else, then the angle isn’t 120º.

So, say for the hour/minute pair, the angle is 120º if and only if:

(M(t) – H(t))t = k +1/3 OR (M(t) – H(t))t = k +2/3, where k is an integer

We can rearrange these as:

t = (k +1/3)/(11/12) OR t = (k +2/3)/(11/12)

We can then just plug in all integers and output a list of valid times t. (We don’t need to do all integers, since t really is a number of at least 0 and less than 12. We start with k=0, and increase it until the resulting t is greater than 12.)

This generates a list of 22 times in a full cycle when the hour and minute hand are exactly 120º apart.

We can do the same with the minute and second hands. We get a list of 1417 times when they are exactly 120º apart. We then compare our two lists looking for times when both of these pairs are 120º apart.

But no values appear in both lists. Therefore there is no time then the hands are truly “equal distances apart”.

So to find an answer to this puzzle, we need to remove at least one of our two assumptions. I first tried removing the “exactly 120º” assumption, and instead went hunting for times when the hands would look equally spaced apart to the naked eye. After all, the puzzle did say “appear equal distances apart”.

For any two pairs of hands, I made a function that is zero if the angle is exactly 120º, and a positive number if not, specifically being a larger number the further we get from 120º. For, say, the minute/hour pair, the function is:

(where |x| is absolute value of x and {x} is the fractional part of x)

This function works as follows: Take the difference in total revolutions at time t. Discard the integer component and just keep the fractional part. For the angle to be 120º, then this figure is either 1/3 or 2/3; either way it is “1/6 away from 1/2, though in either direction”. The function gives zero when the angle is exactly 120º, and increases as the angle drifts from this goal up to a maximum output of 1/3.

I defined “distance off perfect” as being the largest of the three of these functions.

For every time between zero and twelve, we can find the worst result of this function for the three pairs. This gives the following graph:

I zoom in the y-axis so we can see how close those spikes get to zero:

I can then read points off the graph that are close to zero, and those will be close to ‘equal distance’.

The next times of day after 2:55:35 that might look good are:

• 3 hours, 37 minutes, 58.15 seconds after midnight. Positions are 348.91º, 227.82º, 108.98º; angles are 118.83º, 121.10º, and 120.07º. (So decent, but not within 1 degree.)
• 5 hours, 49 minutes, 9.16 seconds after midnight. Positions are 54.94º, 294.92º, 174.58º; angles are 119.64º, 120.34º, and 120.02º.
• 9 hours, 5 minutes, 25.45 seconds after midnight. Positions are 152.54º, 32.54º, 272.71º; angles are 120º exactly, 120.17º, and 119.83º.

The last position gives exactly the same angles as the example image in the puzzle. This is because it is exactly the same amount of time before twelve as the example in the question is after 12, and this produces a mirror image.

I don’t have as much to say about removing the assumption about the hand moving continuously, as I don’t know enough about how clocks work to give a full answer. But I think it doesn’t help us find a perfect moment of equal distances. My understanding of a ticking clock mechanism is that it will jump forward every second but it will never produce a position that was unseen from those in the continuous model. You’d need some sort of clock where each hand is jumping on a different mechanism?

Overall, if I was submitting an answer to the competition, I’d submit 5:49:09. If the organisers noted the hands weren’t perfectly equal distances apart, I’d rebut that neither were they in their own example.

(Thanks, Alex. I wonder now whether Loyd had in mind some “trick” solution such as noon or midnight.)