“A Match Puzzle”

This puzzle, by T.E. Maw of the Luton Public Library, appeared in the Strand in April 1911:

“Take fifteen matches and place them as shown in the diagram, then take away three, change the position of one, and the result will be the word showing what matches are made of.”

	Select Click for Answer

September 9, 2021September 8, 2021 | Puzzles

Exam Week

A problem submitted by the United States and shortlisted for the 16th International Mathematical Olympiad, Erfurt-Berlin, July 1974:

Alice, Betty, and Carol took the same series of examinations. There was one grade of A, one grade of B, and one grade of C for each examination, where A, B, C are different positive integers. The final test scores were

Alice: 20
Betty: 10
Carol: 9

If Betty placed first in the arithmetic examination, who placed second in the spelling examination?

	Select Click for Answer
Call the number of exams n. Then the total number of points distributed is n(A + B + C) = 20 + 10 + 9 = 39. Since A, B, and C are distinct, their sum must be at least 1 + 2 + 3 = 6, and the only combination of factors that can give us 39 is n = 3 and A + B + C = 13. Assume without loss of generality that A > B > C. We’re told that Betty placed first in arithmetic, so she received A points on that exam. But we also know that she received 10 points overall, which is fewer than A + B + C = 13, so she must have received C points in both remaining exams (one of which is spelling). Carol received 9 points overall, which is also less than A + B + C = 13. But she must have received at least B points on two of the exams (since Betty accounts for two Cs), one of which is spelling. If she’d received A points in the spelling exam, though, she’d have a total of at least A + B + C = 13. That’s a contradiction. So Carol must have scored B points in spelling, placing second. (It can be worked out that Alice, Betty, and Carol scored B + A + A, A + C + C, and C + B + B, respectively, and that A = 8, B = 4, and C = 1.) (From Dušan Djukic et al., The IMO Compendium: A Collection of Problems Suggested for the International Mathematical Olympiads: 1959-2004, 2006.)

Click for Answer

September 8, 2021September 7, 2021 | Puzzles

To Town

A problem from the 1949 problems drive of the Archimedeans, the mathematical society of Cambridge University:

A farmer lives in a cottage 4/17 of a mile from a main road. There is a lane leading from his farm to the nearest point Q on the road. The road is straight running north and south, and there is a village two miles south of Q at which he keeps a bicycle. He wishes to go to a town on the road four miles north of Q. He can walk across the fields surrounding the roads at 1 1/2 miles per hour, but along the roads he can walk at 3 1/2 miles per hour. He can cycle at 14 miles per hour. Should he collect his bicycle in order to get to the town from his farm as quickly as possible?

	Select Click for Answer
He should collect the bicycle. By symmetry he can reach point P two miles north of Q in the same time as it takes to reach the village. He can cycle from the village to the town quicker than he can walk there from P. (From the October 1949 issue of the society’s journal, Eureka. The Eureka archive is published under a Creative Commons license.)

September 7, 2021September 6, 2021 | Puzzles

$50 for All

A puzzle by Ben H., a systems engineer at the National Security Agency, from the agency’s August 2016 Puzzle Periodical:

At a work picnic, Todd announces a challenge to his coworkers. Bruce and Ava are selected to play first. Todd places $100 on a table and explains the game. Bruce and Ava will each draw a random card from a standard 52-card deck. Each will hold that card to his/her forehead for the other person to see, but neither can see his/her own card. The players may not communicate in any way. Bruce and Ava will each write down a guess for the color of his/her own card, i.e. red or black. If either one of them guesses correctly, they both win $50. If they are both incorrect, they lose. He gives Bruce and Ava five minutes to devise a strategy beforehand by which they can guarantee that they each walk away with the $50.

Bruce and Ava complete their game and Todd announces the second level of the game. He places $200 on the table. He tells four of his coworkers — Emily, Charles, Doug and Fran — that they will play the same game, except this time guessing the suit of their own card, i.e. clubs, hearts, diamonds or spades. Again, Todd has the four players draw cards and place them on their foreheads so that each player can see the other three players’ cards, but not his/her own. Each player writes down a guess for the suit of his/her own card. If at least one of them guesses correctly, they each win $50. There is no communication while the game is in progress, but they have five minutes to devise a strategy beforehand by which they can be guaranteed to walk away with $50 each.

For each level of play — 2 players or 4 players — how can the players ensure that someone in the group always guesses correctly?

	Select Click for Answer
The 2-person case is simpler. Let r denote a red card and b denote a black card. Let’s list Bruce’s card’s color first. Then the 4 distinct possibilities are rr, bb, rb, and br. The following strategy is foolproof. Bruce guesses that his card is the same color as Ava’s (covering the cases rr and bb). Ava takes the opposite approach and assumes that her card’s color is different from Bruce’s (covering the cases rb and br). One of them will necessarily be correct. If Bruce and Ava properly employ these strategies, they cannot lose the game. Before diving into the 4-person case, let’s reexamine the 2-person case. We could interpret a red card as a 0 and a black card as a 1. Then Bruce’s strategy, as described above, handles the cases 00 and 11. His strategy could be interpreted as “the sum of our cards is even.” (0+0=0, 1+1=2) In this language, Ava’s strategy would be “the sum of our cards is odd.” Again, one of them must be right, no matter the cards. Note that Bruce and Ava have divided the sample space of all possible outcomes into 2 cases, based on the sum of the cards. The 4-person case is more general and depends upon the notion of modular arithmetic, and in particular remainders when dividing by 4. Rather than purely guessing “the suits are the same,” or “the suits are different,” the players must be more clever. Let’s map each suit to a number. Clubs are 0, diamonds are 1, hearts are 2, and spades are 3. So, if Emily looks out and sees hearts (Charles), hearts (Doug), and spades (Fran), she interprets this as 2+2+3=7. Suppose that Emily’s card’s suit is clubs, which she obviously doesn’t know. Consider the following strategies: Emily guesses her suit so that the total sum is a multiple of 4. She sees hearts, hearts, spades and guesses diamonds (2+2+3+1=8). Charles guesses his suit so that the total sum is 1 more than a multiple of 4. He sees clubs, hearts, spades and guesses clubs (0+2+3+0=5). Doug guesses his suit so that the total sum is 2 more than a multiple of 4. He sees clubs, hearts, spades and guesses diamonds (0+2+3+1=6). Fran guesses her suit so that the total sum is 3 more than a multiple of 4. She sees clubs, hearts, hearts and guesses spades (0+2+2+3=7). In this example, Fran guesses correctly because the total sum is 7, which is 3 more than a multiple of 4. For any combination of suits, one of them must be right. After all, the total sum must be a whole number. And all whole numbers have remainder 0, 1, 2, or 3 when divided by 4. As long as each of them sticks to his/her strategy, the group can ensure they won’t lose the game. Interestingly, it doesn’t matter which numbers they choose to use, as long as they all have different remainders when divided by 4. So, choosing any four consecutive whole numbers is a sufficient mapping.

Click for Answer

The 2-person case is simpler. Let r denote a red card and b denote a black card. Let’s list Bruce’s card’s color first. Then the 4 distinct possibilities are rr, bb, rb, and br. The following strategy is foolproof. Bruce guesses that his card is the same color as Ava’s (covering the cases rr and bb). Ava takes the opposite approach and assumes that her card’s color is different from Bruce’s (covering the cases rb and br). One of them will necessarily be correct. If Bruce and Ava properly employ these strategies, they cannot lose the game.

Before diving into the 4-person case, let’s reexamine the 2-person case. We could interpret a red card as a 0 and a black card as a 1. Then Bruce’s strategy, as described above, handles the cases 00 and 11. His strategy could be interpreted as “the sum of our cards is even.” (0+0=0, 1+1=2) In this language, Ava’s strategy would be “the sum of our cards is odd.” Again, one of them must be right, no matter the cards. Note that Bruce and Ava have divided the sample space of all possible outcomes into 2 cases, based on the sum of the cards.

The 4-person case is more general and depends upon the notion of modular arithmetic, and in particular remainders when dividing by 4. Rather than purely guessing “the suits are the same,” or “the suits are different,” the players must be more clever. Let’s map each suit to a number. Clubs are 0, diamonds are 1, hearts are 2, and spades are 3. So, if Emily looks out and sees hearts (Charles), hearts (Doug), and spades (Fran), she interprets this as 2+2+3=7. Suppose that Emily’s card’s suit is clubs, which she obviously doesn’t know. Consider the following strategies:

Emily guesses her suit so that the total sum is a multiple of 4. She sees hearts, hearts, spades and guesses diamonds (2+2+3+1=8).
Charles guesses his suit so that the total sum is 1 more than a multiple of 4. He sees clubs, hearts, spades and guesses clubs (0+2+3+0=5).
Doug guesses his suit so that the total sum is 2 more than a multiple of 4. He sees clubs, hearts, spades and guesses diamonds (0+2+3+1=6).
Fran guesses her suit so that the total sum is 3 more than a multiple of 4. She sees clubs, hearts, hearts and guesses spades (0+2+2+3=7).

In this example, Fran guesses correctly because the total sum is 7, which is 3 more than a multiple of 4. For any combination of suits, one of them must be right. After all, the total sum must be a whole number. And all whole numbers have remainder 0, 1, 2, or 3 when divided by 4. As long as each of them sticks to his/her strategy, the group can ensure they won’t lose the game. Interestingly, it doesn’t matter which numbers they choose to use, as long as they all have different remainders when divided by 4. So, choosing any four consecutive whole numbers is a sufficient mapping.

August 27, 2021August 26, 2021 | Puzzles

Paint Job

chessboard

Given a standard chessboard, you can choose any rank or file and repaint each of its squares to the opposite color (white squares turn black, and black squares turn white). By doing this repeatedly, is it possible to produce a board with 63 white squares and one black square?

	Select Click for Answer
No. Repainting a rank or file in this way transforms its k black and (8 – k) white squares into (8 – k) black and k white squares. The number of black squares changes by (8 – k) – k = 8 – 2k. This is an even number, which means that the parity of the number of black squares is preserved. Since we start with an even number (32) of black squares, there is no way to reach the odd outcome 1. (From Viktor Prasolov’s Problems in Plane and Solid Geometry, 1988.)

Click for Answer

August 26, 2021August 25, 2021 | Puzzles

Black and White

cumming chess problem

A tidy problem by G.A.W. Cumming. White to mate in two moves.

	Select Click for Answer
1. Kxc6. From Noah’s Sunday Times, via Charles F. Wadworth’s Unique Chess Problems, 1890.

August 23, 2021August 21, 2021 | Puzzles

The Picnic

We’re giving out apples to a group of boys. If we distribute the entire supply, then every boy will get three, except for one, who will get two. If instead we give each boy two apples, then we’ll have eight apples left over. How many apples are there altogether?

	Select Click for Answer
This can be done intuitively. Imagine handing out the apples as described in the first case and then taking back one apple from each of the boys who received three. At that point we’ll be holding eight apples. So there are 9 boys and 26 apples.

August 13, 2021August 13, 2021 | Puzzles

Podcast Episode 353: Lateral Thinking Puzzles

Here are six new lateral thinking puzzles — play along with us as we try to untangle some perplexing situations using yes-or-no questions.

Intro:

Lili McGrath’s 1915 “floor polisher” is a pair of slippers connected by a cord.

Eighteenth-century English landowners commissioned custom ruins.

The sources for this week’s puzzles are below. In some cases we’ve included links to further information — these contain spoilers, so don’t click until you’ve listened to the episode:

Puzzle #1 is from listener Moxie LaBouche.

Puzzle #2 is from listener Cheryl Jensen, who sent this link.

Puzzle #3 is from listener Theodore Warner. Here’s a link.

Puzzle #4 is from listener David Morgan.

Puzzle #5 is from listener Bryan Ford, who sent these links.

Puzzle #6 is from listener John Rusk, who sent this link.

You can listen using the player above, download this episode directly, or subscribe on Google Podcasts, on Apple Podcasts, or via the RSS feed at https://futilitycloset.libsyn.com/rss.

Please consider becoming a patron of Futility Closet — you can choose the amount you want to pledge, and we’ve set up some rewards to help thank you for your support. You can also make a one-time donation on the Support Us page of the Futility Closet website.

Many thanks to Doug Ross for the music in this episode.

If you have any questions or comments you can reach us at podcast@futilitycloset.com. Thanks for listening!

August 9, 2021August 8, 2021 | Podcast · Puzzles

Skulduggery

This secret message appears in J.J. Connington’s 1933 novel Tom Tiddler’s Island:

TEIIL LFILH TCETU FDHSO OENPR YYUGO HNGOF
LOVTU GCHAN NOATN AEHAT ISUWE ETFST GSCAD
OFRGH PELPE HASLE GASTH HGSMR LHLAR ARNIF
THRDL NITFO SSWSG NYILE EFALT ODECT IESOL
NTSNT COOUE AODNT IUTAI TIOON LEANR IIGOT
AHNOM FINHE YLMFD ATTTS MANHH OFEII ETODD
OTPCA MOTIE FMONG IMCLA TTCHB YIMNN ETROX
EMCOU VSFHE ELMPN NCTAW ETRWO OAHEE IYCNA
OIRBT RTXET PEIZN RSCSA TIKOH NITHT EMFNE
NNRUO GOTGP ENETP SYANS Z

What does it mean?

	Select Click for Answer
It’s a transposition cipher. Lay out the 336 letters into 4 rows of 84 characters: TEIILLFILHTCETUFDHSOOENPRYYUGOHNGOFLOVTUGCHANNOATNAEHATISUWEETFSTGSCADOFRGHPELPEHASL EGASTHNGSMRLHLARARNIFTHRDLNITFOSSWSGNYILEEFALTODECTIESOLNTSNTCOOUEAODNTIUTAITIOONLEA NRIIGOTAHNOMRINHEYLMFDSTTTSMANHHOFEIIETODDOTPCAMOTIEFMONGIMCLATTCHBYIMNNETROXEMCOUVS FHAELMPNNCTAWETRWOOAHEEIYCNAOIRBTRTXETPEIZNRSCSATIKOHNITHTEMFNENNRUOGOTGPENETPSYANSZ Cut that in two and lay the two halves alongside one another: TEIILLFILHTCETUFDHSOOENPRYYUGOHNGOFLOVTUGC EGASTHNGSMRLHLARARNIFTHRDLNITFOSSWSGNYILEE NRIIGOTAHNOMRINHEYLMFDSTTTSMANHHOFEIIETODD FHAELMPNNCTAWETRWOOAHEEIYCNAOIRBTRTXETPEIZ HANNOATNAEHATISUWEETFSTGSCADOFRGHPELPEHASL FALTODECTIESOLNTSNTCOOUEAODNTIUTAITIOONLEA OTPCAMOTIEFMONGIMCLATTCHBYIMNNETROXEMCOUVS NRSCSATIKOHNITHTEMFNENNRUOGOTGPENETPSYANSZ Interleave the rows: TEIILLFILHTCETUFDHSOOENPRYYUGOHNGOFLOVTUGC HANNOATNAEHATISUWEETFSTGSCADOFRGHPELPEHASL EGASTHNGSMRLHLARARNIFTHRDLNITFOSSWSGNYILEE FALTODECTIESOLNTSNTCOOUEAODNTIUTAITIOONLEA NRIIGOTAHNOMRINHEYLMFDSTTTSMANHHOFEIIETODD OTPCAMOTIEFMONGIMCLATTCHBYIMNNETROXEMCOUVS FHAELMPNNCTAWETRWOOAHEEIYCNAOIRBTRTXETPEIZ NRSCSATIKOHNITHTEMFNENNRUOGOTGPENETPSYANSZ Combine the halves again and read first the odd and then the even-numbered columns: TEIILLFILHTCETUFDHSOOENPRYYUGOHNGOFLOVTUGCNRIIGOTAHNOMRINHEYLMFDSTTTSMANHHOFEIIETODD HANNOATNAEHATISUWEETFSTGSCADOFRGHPELPEHASLOTPCAMOTIEFMONGIMCLATTCHBYIMNNETROXEMCOUVS EGASTNHGSMRLHLARARNIFTHRDLNITFOSSWSGNYILEEFHAELMPNNCTAWETRWOOAHEEIYCNAOIRBTRTXETPEIZ FALTODECTIESOLNTSNTCOOUEAODNTIUTAITIOONLEANRSCSATIKOHNITHTEMFNENNRUOGOTGPENETPSYANSZ “THE FINAL LOT OF THE LAST THREE THOUSAND WAS SENT OFF ON THURSDAY AND GOT THROUGH SAFE STOP NOTHING SEEN OF NIPASGAL STOP THINK OF THROWING THEM WELL OFF THE SCENT BY USING ANOTHER PORT NEXT TIME STOP ADVISE AGAINST LANDING CHEMICALS TILL FURTHER NOTICE STOP GREY CLOUD IN OFFING STOP WIL GIVE YOU ALL CLEAR THRICE COMMA AT NINE COMMA NINE THIRTY COMMA AND TEN THIRTY COMMA ON NIGHT BEFORE I EXPECT YOU ENDS ZZ” The method of solution is laid out on pages 148-161 of the novel. (Via Craig Bauer’s 2016 book Secret History: The Story of Cryptology.) 08/21/2021 UPDATE: Connington’s text contains six errors, which are carried through in Bauer’s exercise. I’d caught two of these in writing this post but overlooked the other four. Reader Peter Dawyndt kindly sent the ciphertext as it should have appeared: TEIIL LFILH TCETU FDHSO OENPR YYUGO HNGOF LOVTU GCHAN NOATN AEHAT ISUWE ETFST GSCAD OFRGH PELPE HASLE GASTN HGSMR LHLAR ARNIF THRDL NITFO SSWSG NYILE EFALT ODECT IESOL NTSNT COOUE AODNT IUTAI TIOON LEANR IIGOT AHNOM RINHE YLMFD STTTS MANHH OFEII ETODD OTPCA MOTIE FMONG IMCLA TTCHB YIMNN ETROX EMCOU VSFHA ELMPN NCTAW ETRWO OAHEE IYCNA OIRBT RTXET PEIZN RSCSA TIKOH NITHT EMFNE NNRUO GOTGP ENETP SYANS Z Also, I introduced one typo in the plaintext — WIL should be WILL. Sorry for these errors; this cipher must have frustrated Connington’s readers. (Thanks, Peter.)

Click for Answer

It’s a transposition cipher. Lay out the 336 letters into 4 rows of 84 characters:

TEIILLFILHTCETUFDHSOOENPRYYUGOHNGOFLOVTUGCHANNOATNAEHATISUWEETFSTGSCADOFRGHPELPEHASL
EGASTHNGSMRLHLARARNIFTHRDLNITFOSSWSGNYILEEFALTODECTIESOLNTSNTCOOUEAODNTIUTAITIOONLEA
NRIIGOTAHNOMRINHEYLMFDSTTTSMANHHOFEIIETODDOTPCAMOTIEFMONGIMCLATTCHBYIMNNETROXEMCOUVS
FHAELMPNNCTAWETRWOOAHEEIYCNAOIRBTRTXETPEIZNRSCSATIKOHNITHTEMFNENNRUOGOTGPENETPSYANSZ

Cut that in two and lay the two halves alongside one another:

TEIILLFILHTCETUFDHSOOENPRYYUGOHNGOFLOVTUGC
EGASTHNGSMRLHLARARNIFTHRDLNITFOSSWSGNYILEE
NRIIGOTAHNOMRINHEYLMFDSTTTSMANHHOFEIIETODD
FHAELMPNNCTAWETRWOOAHEEIYCNAOIRBTRTXETPEIZ

HANNOATNAEHATISUWEETFSTGSCADOFRGHPELPEHASL
FALTODECTIESOLNTSNTCOOUEAODNTIUTAITIOONLEA
OTPCAMOTIEFMONGIMCLATTCHBYIMNNETROXEMCOUVS
NRSCSATIKOHNITHTEMFNENNRUOGOTGPENETPSYANSZ

Interleave the rows:

TEIILLFILHTCETUFDHSOOENPRYYUGOHNGOFLOVTUGC
HANNOATNAEHATISUWEETFSTGSCADOFRGHPELPEHASL
EGASTHNGSMRLHLARARNIFTHRDLNITFOSSWSGNYILEE
FALTODECTIESOLNTSNTCOOUEAODNTIUTAITIOONLEA

NRIIGOTAHNOMRINHEYLMFDSTTTSMANHHOFEIIETODD
OTPCAMOTIEFMONGIMCLATTCHBYIMNNETROXEMCOUVS
FHAELMPNNCTAWETRWOOAHEEIYCNAOIRBTRTXETPEIZ
NRSCSATIKOHNITHTEMFNENNRUOGOTGPENETPSYANSZ

Combine the halves again and read first the odd and then the even-numbered columns:

TEIILLFILHTCETUFDHSOOENPRYYUGOHNGOFLOVTUGCNRIIGOTAHNOMRINHEYLMFDSTTTSMANHHOFEIIETODD
HANNOATNAEHATISUWEETFSTGSCADOFRGHPELPEHASLOTPCAMOTIEFMONGIMCLATTCHBYIMNNETROXEMCOUVS
EGASTNHGSMRLHLARARNIFTHRDLNITFOSSWSGNYILEEFHAELMPNNCTAWETRWOOAHEEIYCNAOIRBTRTXETPEIZ
FALTODECTIESOLNTSNTCOOUEAODNTIUTAITIOONLEANRSCSATIKOHNITHTEMFNENNRUOGOTGPENETPSYANSZ

“THE FINAL LOT OF THE LAST THREE THOUSAND WAS SENT OFF ON THURSDAY AND GOT THROUGH SAFE STOP NOTHING SEEN OF NIPASGAL STOP THINK OF THROWING THEM WELL OFF THE SCENT BY USING ANOTHER PORT NEXT TIME STOP ADVISE AGAINST LANDING CHEMICALS TILL FURTHER NOTICE STOP GREY CLOUD IN OFFING STOP WIL GIVE YOU ALL CLEAR THRICE COMMA AT NINE COMMA NINE THIRTY COMMA AND TEN THIRTY COMMA ON NIGHT BEFORE I EXPECT YOU ENDS ZZ”

The method of solution is laid out on pages 148-161 of the novel.

(Via Craig Bauer’s 2016 book Secret History: The Story of Cryptology.)

08/21/2021 UPDATE: Connington’s text contains six errors, which are carried through in Bauer’s exercise. I’d caught two of these in writing this post but overlooked the other four. Reader Peter Dawyndt kindly sent the ciphertext as it should have appeared:

TEIIL LFILH TCETU FDHSO OENPR YYUGO HNGOF
LOVTU GCHAN NOATN AEHAT ISUWE ETFST GSCAD
OFRGH PELPE HASLE GASTN HGSMR LHLAR ARNIF
THRDL NITFO SSWSG NYILE EFALT ODECT IESOL
NTSNT COOUE AODNT IUTAI TIOON LEANR IIGOT
AHNOM RINHE YLMFD STTTS MANHH OFEII ETODD
OTPCA MOTIE FMONG IMCLA TTCHB YIMNN ETROX
EMCOU VSFHA ELMPN NCTAW ETRWO OAHEE IYCNA
OIRBT RTXET PEIZN RSCSA TIKOH NITHT EMFNE
NNRUO GOTGP ENETP SYANS Z

Also, I introduced one typo in the plaintext — WIL should be WILL. Sorry for these errors; this cipher must have frustrated Connington’s readers. (Thanks, Peter.)