No. The player with the Xs has won and so must have made the last move. If that player moved first, then there ought to be one more X than there are Os in the diagram. If the player making Os went first, then there should be an equal number of Os and Xs. The diagram contains more Os than Xs, so the position is impossible.

From Dr. Crypton and His Problems: Mind Benders From Science Digest, 1982.

Bert is 72, Ben is 80, and Bill is 85.

Start both timers. When the first one reaches its limit, turn it over. Do the same with the second. When the first one again reaches its limit, 8 minutes altogether have passed, which means that the second timer has been running for 1 minute since you restarted it. Invert that timer, and when the minute’s worth of sand has run out, 9 minutes altogether will have passed.

Similar: How can you measure 45 minutes using two fitful hour fuses?

Consider three groups of people: N contains unsociable weirdos, W contains all the other weirdos, and U contains all the non-weird unsociable people. Let n, w, and u be the number of people in each of these groups, correspondingly, and let a be the number of pairs of acquaintances, each containing one person from W and another from U. We need to prove that w + n < u + n, or w < u.

Well, no one in N can have an acquaintance in W, since weirdos don’t associate with sociable people. Since acquaintanceship is reciprocal, this means that each of the people in W is finding all their numerous (10+) friends in U. So a is at least 10w.

A person in U can find friends anywhere but must have fewer than 10 of them. So a is less than 10u.

Combining these inferences, if 10w ≤ a and a < 10u, then 10w < 10u and w < u.

11/03/2025 UPDATE: Reader Wolfgang Schilhan writes:

Hi! I noticed there is a flaw in the puzzle “A Prickly Puzzle” from October 24.

While the number of weirdos can’t be larger than the number of unsociable people, it still can be equal.

Consider a group of 10 or fewer people, each of them mutually acquainted but with no other acquaintances outside this group. Then everyone in this group is unsociable, since they have less than 10 aquaintances, and also a weirdo, since all their acquaintances are unsociable. Therefore w = u.

Imagine a society consisting only of such groups, plus any number of sociable people. Or a network of groups of 9 or lesser people like above, where anyone can have at most one acquaintance from another group. In all those examples, every unsociable person is also a weirdo and vice versa, therefore the number of weirdos equals the number of unsociable people.

The solution ignores the cases where the groups W and U are empty.

The error in the proof from the solution lies in the line “A person in U can find friends anywhere but must have fewer than 10 of them. So a is less than 10u.” This is true only if u > 0.

If u = 0, then a also is zero and therefore not less than, but equal to 10u.

(Thanks, Wolfgang.)

At first it seems we have too little information, but it turns out the angle is invariant with the size of the larger square — grab point H and drag it left and right:

(GeoGebra resource by John Golden.)

The ball is always working against air resistance, so it’s losing energy all the time; at a given height, its total energy while rising is greater than while falling. The ball’s potential energy at this height is the same in both cases, so its kinetic energy (and hence its speed) is greater when it’s rising. So it takes longer to fall than to rise.

1. Kf5! Now the bishop must move, and then 2. h8=Q+ Bb8 3. Qh1#

Note that the king’s waiting move must be to f5. Moving to a black square leaves him vulnerable to a check; moving to the h-file blocks the queen’s access to h1; and moving to f3 blocks the queen’s final checkmate.

If 1. Kf5 Be5 then 2. h8=Q+ Bxh8 3. f8=Q#.

From Eskilstuna-Kuriren, 1931.

“If coal is burnt on the third floor height, the potential energy of the combustion products (water, ashes, carbon dioxide, carbon monoxide, and unburnt coal particles) will increase exactly as much as the coal potential energy.”

From Lange’s 1987 book Physical Paradoxes and Sophisms.

If all four equations are satisfied, then multiplying all four equations together will produce another valid equation. We obtain:

a⁷ × b⁶ × c⁵ × d⁴ × e³ × f² × g = 2,677,277,333,530,800,000.

The number on the right factors into prime powers as 2,677,277,333,530,800,000 = 2⁷ × 3⁶ × 5⁵ × 7⁴ × 11³ × 13² × 17, as can be determined by trial division by the small primes 2, 3, 5, 7, 11, 13, and 17, for instance. So we have the new equation:

a⁷ × b⁶ × c⁵ × d⁴ × e³ × f² × g = 2⁷ × 3⁶ × 5⁵ × 7⁴ × 11³ × 13² × 17.

If p is any prime dividing a, then p⁷ divides the right hand side. Since only 2 appears to the seventh power, the only value p can take is 2. Since a must be greater than 1, we see that 2 must divide a. Because 2 appears to exactly the seventh power on the right, we see that a = 2. Canceling a⁷ on the left with 2⁷ on the right leaves:

b⁶ × c⁵ × d⁴ × e³ × f² × g = 3⁶ × 5⁵ × 7⁴ × 11³ × 13² × 17.

Repeating the same argument shows that b = 3, c = 5, d = 7, e = 11, f = 13, g = 17 is the unique solution to the new equation. However, it is still necessary that these values actually satisfy the original four equations, which in fact they do as can be checked easily. (To see why this last step is necessary, try using the same reasoning on the equations: x = 10, y² = 9, z³ = 25, which clearly has no solutions in integers.)

10/20/2025 UPDATE: A number of readers point out that this yields pretty readily to standard solution. Simon Loria writes:

If you factor Eq 1 (a relatively easy task) you get 2² × 3 × 5² x 17. Even if you don’t identify the sequence at this point (17 being the 7th (gth) simple prime), knowing a & c allows you to confirm a = 2 & c = 5 and solve for d in Eq 3. This in turn allows you to solve for e in Eq 4 and f in Eq 2.

And Catalin Voinescu writes:

The fourth equation isn’t even necessary. It makes the clever solution
possible, but you can approach this more directly and get there quicker,
with arithmetic that fits on a pocket calculator.

a² * b * c² * g = 5100 = 2² * 3 * 5² * 17

The right-hand side can be written as a product of at most six factors
that are greater than 1, and the left-hand side is a product of six
factors that are greater than 1, so each factor on the left corresponds
to one on the right. The repeated ones are a and c on the left and 2 and
5 on the right, so a and c must be 2 and 5 (not necessarily in this
order); b and g must be 3 and 17 (again, not necessarily in this order).

a * b² * e * f² = 33462 = 2 * 3² * 11 * 13²

Same logic tells us a and e are 2 and 11, and b and f are 3 and 13.

Using both results, a = 2, so c = 5, e = 11; and b = 3, thus g = 17, f = 13.

Finally, from

a * c² * d³ = 17150 = 2 * 5² * 7³

we immediately get d = 7, using the same argument, or even simply
plugging in a = 2, c = 5 and solving for d. All seven values are now
known. The fourth equation is redundant.

Even if you follow the given solution, if you multiply the numbers and
only then decompose the large number into prime factors, you’re doing
way more work than if you factor the four smaller numbers first…

(Thanks to both.)

From George Arnold et al., The Magician’s Own Book, 1857.

10/10/2025 UPDATE: The published solution works, but there’s an alternative that gets the job done in 10 steps:

Start                 12  0  0
Fill 7 from 12         5  7  0
Fill 5 from 7          5  2  5
Empty 5 into 12       10  2  0
Empty 7 into 5        10  0  2
Fill 7 from 12         3  7  2
Fill 5 from 7          3  4  5
Empty 5 into 12        8  4  0
Empty 7 into 5         8  0  4
Fill 7 from 12         1  7  4
Fill 5 from 7          1  6  5

That’s from reader Catalin Voinescu. Many thanks to him and to everyone else who wrote in.

10/12/2025 UPDATE: Reader Chris Hills presents a standard way of solving this type of problem:

In this diagram,

each point represents a state of how much wine is in each bottle, the vertical distance from the point to the bottom edge of the big equilateral triangle is the number of pints in the 5-pint bottle, the vertical distance to the left edge is the number in the 7-pint bottle, and the vertical distance to the right edge is the number in the 12-pint bottle. These 3 distances always add up to 12 because of Viviani’s theorem, which you mention in “The Glass Rod Problem.”

The amount of wine in each bottle is ≥ 0 pints and ≤ its capacity, so only points in the green area are possible. The points outside the green area represent situations where a bottle is holding more than its capacity.

A transfer from one bottle to another is represented by a line from one edge of the green parallelogram to another edge. We can’t stop in the middle, we can only stop a transfer when the giving bottle is empty or the receiving bottle is full.

The starting position is the bottom left corner, where all the wine is in the 12-pint bottle. The end position is anywhere on the black line, this is where there are 6 pints in the 7-pint bottle. In this sort of problem, there are always either no solutions or exactly 2 solutions. The solution you gave follows the blue arrows, and takes 12 steps. The red arrows give the other solution, which takes 10 steps. [Thanks, Chris.]

In a related class of problems, known as water-fetching puzzles, water is drawn from a well and a prescribed amount must be measured using jugs of stated sizes. H.D. Grossman gave a geometric approach in Scripta Mathematica in 1948.