Query

From Gerald Lynton Kaufman’s The Book of Modern Puzzles (1954):

All DROONS are the same size and shape.
All green SLACKENS are the same size and shape.
Twenty DROONS just fill up a MULDRUFF.
All WALLAXES contain green SLACKENS.
A green SLACKEN is 10% bigger than a DROON.
A WALLAX is smaller than a MULDRUFF.

“If all MULDRUFFS and all WALLAXES are predominantly RED throughout, what is the largest possible number of green SLACKENS in a WALLAX?”

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Eight. If a MULDRUFF holds 20 DROONS, and a green SLACKEN is 10% bigger than a DROON, then a MULDRUFF can accommodate at most 18 green SLACKENS. And if a WALLAX is smaller than this, then it can hold at most 17 green SLACKENS. But if each WALLAX is predominantly red, then the proportion of green SLACKENS in its contents can’t be more than one-half. So the largest number of green SLACKENS it can contain is 8. 11/22/2025 UPDATE: This is just wrong. Let a DROON have size 1. Then a MULDRUFF has capacity 20, and a green SLACKEN has size 1.1. So our MULDRUFF will accommodate 9 green SLACKENs (= 9.9) and 10 DROONS (= 19.9 < 20). Now we can transfer this cargo to a WALLAX of, say, capacity 19.95 and fulfill the terms of the problem with 9 (not 8) green SLACKENS — the WALLAX remains predominantly red by both number and volume. Thanks to everyone who wrote in about this.

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November 20, 2025December 13, 2025 | Puzzles

Constitution

A sobering problem from Gerald Lynton Kaufman’s Book of Modern Puzzles, 1954:

If a GLEEPER is as long as two PLONTHS and a half-GLEEPER, and a BLAHMIE is as long as two GLEEPERS and a half-BLAHMIE, and a POOSTER is as long as two BLAHMIES and a half-POOSTER, then how many PLONTHS long is a half-POOSTER?

“It may help you to make a sketch.”

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Four PLONTHS make a GLEEPER, four GLEEPERS make a BLAHMIE, and four BLAHMIES make a POOSTER. So half a POOSTER is 32 PLONTHS.

November 10, 2025November 9, 2025 | Puzzles

Black and White

brotherton chess problem

With this problem D.W.A. Brotherton won second prize in a 1955 competition for British composers under 21. White to mate in two moves.

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1. Rf5! 1. … Kh7+ 2. Rf8# 1. … Rh7 2. Rb8# 1. … g6/5 2. Bc4# 1. … B any 2. Rxg7# From The Problemist.

November 6, 2025November 5, 2025 | Puzzles

Character Study

A puzzle by Paul Hoffman, from Science Digest. Could this game ever have resulted from a strict adherence to the rules of tic-tac-toe (noughts and crosses)?

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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.

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November 3, 2025November 2, 2025 | Puzzles

Relative

A problem from Joseph Madachy’s Mathematics on Vacation (1966):

When Bert was just one year younger than Bill was when Ben was half as old as Bill will be 3 years from now, Ben was twice as old as Bill was when Ben was 1/3 as old as Bert was 3 years ago. But, when Bill was twice as old as Bert, Ben was 1/4 as old as Bill was one year ago.

“Ignoring odd months and considering that Bert has passed the half-century mark, it will be no problem to find out how old these three friends are.”

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

November 2, 2025November 1, 2025 | Puzzles

The Sands of Time

From Howard Dinesman’s Superior Mathematical Puzzles (2003):

How can you measure 9 minutes using two hourglass-style timers, one that measures 4 minutes and the other 7 minutes?

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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?

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October 28, 2025 | Puzzles

A Prickly Puzzle

A problem by F. Nazarov, from the November/December 1994 issue of Quantum:

A person with fewer than 10 acquaintances is unsociable. If all your acquaintances are unsociable, you’re a weirdo. If all acquaintanceships are reciprocal (that is, if you know me then I know you), prove that unsociable people outnumber weirdos.

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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.)

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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.)

October 24, 2025November 3, 2025 | Puzzles

Square Deal

From Catriona Agg’s endlessly rewarding Twitter feed:

The three smaller squares are the same size. What’s the angle? (The outermost shape too is a square.)

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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.)

October 17, 2025 | Puzzles

Up and Down

A remarkably simple question by Russian scientist A. Savin, from the September/October 1995 issue of Quantum:

A ping-pong ball is tossed into the air. Will it take longer to go up or to come back down?

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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.

October 16, 2025 | Puzzles

Black and White

clausén chess problem

By Sigurd Clausén. White to mate in three moves.

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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.

October 15, 2025October 14, 2025 | Puzzles