“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.

No. At the start, the number of -1s on the pentagon’s perimeter is 0, an even number. Each step will change two of these numbers, so their quantity will always be even. Our goal would make it odd. So the task is impossible.

From Arthur Engel’s Problem-Solving Strategies, 2003.

This solution is by Titu Zvonaru of Comănești, Romania. The sum of the distances AB, BD, and DE (3 + 1 + 4) equals the distance AE (8), so cities A, B, D, and E lie on the same straight road. The square of distance AC (25) equals the sum of the squares of CD and AD (9 + 16), so by the converse of the Pythagorean theorem CD is perpendicular to AD. This means that CD and DE are the legs of a right triangle, and Pythagoras shows that hypotenuse CE is 5 km.

1. Re2!, waiting:

1. … d5 2. Re6#
1. … Kb5 2. Rc2#
1. … Kd5 2. Bg2#

By Otto Wurzburg, from the Grand Rapids Herald, 1932.

(Thanks, Derek!)

1. Kg7!, waiting:

1. … Ka1 2. Qb1#
1. … Nb5 2. Qg8#
1. … Ne4 2. Bg8#

Note that the king can’t choose any other square. The eighth rank must remain free for the queen; e7 and f6 are vulnerable to knight checks; 1. Ke6 would block the queen’s check in the second variation; and 1. Kg6 would block the bishop’s diagonal in the first and second.