Puzzle #11: T.C. Wentworth’s Problem

By Muriel Bristol (Transcriber) | March 9, 2019

T.C. Wentworth’s problem – how to construct a square containing 20 square inches: draw a base line 2 inches long. Draw perpendicular 4 inches long at one end of the baseline. Now draw hypotenuse from the above lines. This hypotenuse will be one side of a square containing exactly 20 square inches. The square of the base plus the square of the perpendicular equals the square of the hypotenuse. – Milton Mills, N.H. (Boston Globe, March 13, 1902).

[Answer to Puzzle #11 to follow in the next Puzzle]

Solution to Puzzle #10: J.O. Porter’s Cork Problem

Followers of the Boston Globe’s Puzzle Problem column of long ago answered:

G.W. Monegan, North Chelmsford, says the cork costs 5 cents; so say Frank E. Witherell, Walter L. Colburn, E.G. Hayden, L.N. Lewis, S.O. Keep, Rowley; D.T. Jardine, Cambridge; James A., Newton (Boston Globe, December 25, 1901).

For those that do not want to simply take their word: $1.10 = $1.00 + 2X; $0.10 = 2X; $0.05 = X.

If the cork is worth 5¢, and the bottle is worth that plus a dollar, then the bottle is worth $1.05. Taken altogether, the total is $1.10.

Puzzle #10: J.O. Porter’s Cork Problem

By Muriel Bristol (Transcriber) | November 13, 2018

J.O. Porter, Jr., was son of Milton ice industry magnate J.O. Porter. The younger Porter posed the following problem to the Puzzle Problems column of the Boston Globe in December 1901:

I have been reading your puzzle column and am very much interested therein, and now send one that I think will interest your readers.

If a bottle and its cork cost $1.10 and the bottle cost $1 more than the cork, how much did the cork cost?

J.O. Porter, Jr., Marblehead (Boston Globe, December 19, 1901).

The Porters may have been preparing for their next Milton ice harvest when the younger Porter sent in this relatively simple problem.

[Answer to Puzzle #10 to follow in the next Puzzle]

Solution to Puzzle #9: Puzzle #9: How Did They Get There?

The coal, carrot, and scarf came from a snowman. “Nobody put them on the lawn” because they placed them on the snowman. When the snowman melted, they fell to the lawn.

Yes, yes, this one was much too easy. Commenters thought I might not be well. Philo Beddo got there within minutes, with several others following shortly thereafter.

If you have a nice logic puzzle, situated somewhere between those intended for schoolchildren (like this one evidently was) and those intended for MIT physicists (with charts, diagrams, and calculus), drop it in a comment.

Puzzle #9: How Did They Get There?

By Muriel Bristol (Transcriber) | October 23, 2018

Five pieces of coal, a carrot and a scarf are lying on a hill near a remote house. Nobody put them on the lawn, but there is a perfectly logical reason why they should be there. What is it?

[Answer to Puzzle #9 to follow in the next Puzzle]

Solution to Puzzle #8: Something About Mary

Puzzle #8 was a “trick” question, the sort my grandfather liked to tell.

Mary’s mum had four children: April, May, June, and Mary herself, of course. The answer was in the title.

Puzzle #8: Something About Mary

By Muriel Bristol (Transcriber) | October 16, 2018

Mary’s mum has four children.
The first child is named April.
The second child is named May.
The third child is named June.

What is the name of the fourth child?

[Answer to Puzzle #8 to follow in the next Puzzle]

Solution to Puzzle #7: Double Jeopardy Doors

Double Jeopardy Doors had a certain counter-intuitive aspect that made it more difficult than it would first seem.

Ask each robot in turn, “What the other robot would answer if it were asked which door led to freedom.” If the robot is the liar, it will answer falsely that the truthful robot would direct you to the death door. If the robot is truthful, it will answer truthfully that the liar robot would direct you to the death door.

The counter-intuitive part is that, when you use that particular question, the two robots will agree. Whichever door they both indicate, go through the other one.

Puzzle #7: Double Jeopardy Doors

By Muriel Bristol (Transcriber) | October 9, 2018

You are trapped in a room with two doors. One leads to certain death and the other leads to freedom. You don’t know which is which.

There are two robots guarding the doors. They will let you choose one door but upon doing so you must go through it.

You can, however, ask one robot one question. The problem is one robot always tells the truth, the other always lies and you don’t know which is which.

What is the question you ask?

[Answer to Puzzle #7 to follow in the next Puzzle]

Solution to Puzzle #6: Mislabeled Boxes

You know that the box labeled “Oranges and Apples” is labeled incorrectly. Therefore, it must contain either only apples or only oranges. Pick from that box.

Path One: If you pick an apple, relabel that box “Apples.” The box mislabeled “Oranges” must actually contain the apples and oranges. Relabel it “Oranges and Apples.” Finally, by process of elimination, the remaining box that is mislabeled “Apples” must actually contain oranges. Relabel it “Oranges.”

Path Two: If you pick an orange, relabel that box “Oranges.” The box mislabeled “Apples” must actually contain the apples and oranges. Relabel it “Oranges and Apples.” Finally, by process of elimination, the remaining box that is mislabeled “Oranges” must actually contain apples. Relabel it “Apples.”

Apple Computer used this question in interviews for the position of Software Quality Assurance Engineer.

Congratulations to Mike Sylvia for solving this first, with a Path One solution. He gets the glory, such as it is. Those who followed after him are entitled also to a measure of thanks for their effort and satisfaction in their results.

Puzzle #6: Mislabeled Boxes

By Muriel Bristol (Transcriber) | October 2, 2018

There are three boxes. One is labeled “Apples,” another is labeled “Oranges,” and the last one is labeled “Apples and Oranges.”

You know that each is labeled incorrectly. You may ask me to pick one fruit from one box, which you choose.

How can you label the boxes correctly?

[Answer to Puzzle #6 to follow in the next Puzzle]

Solution to Puzzle #5: Smith, Jones and Robinson

We are told that Mr. Robinson lives in Leeds. Therefore, Mr. Robinson does not live anywhere else and none of the other passengers live in Leeds.

The guard lives halfway between Leeds and Sheffield. His nearest neighbor is a passenger who earns three times as much as the guard. Mr. Jones can not be the guard’s neighbor, because his salary is not divisible by three. Neither can Mr. Robinson, as we have seen that he lives in Leeds. Therefore, the guard’s neighbor must be Mr. Smith.

It also follows that Mr. Jones must live in Sheffield, that being the only remaining choice. Since, the guard’s namesake is said to live in Sheffield, it follows that the guard’s name is Jones.

Smith is said to have beaten the fireman at billiards. Therefore, Smith is not the fireman. We have seen that Jones is the guard. Therefore, by process of elimination, Smith must be the engine driver.


Puzzle #5: Smith, Jones and Robinson

By Muriel Bristol (Transcriber) | September 25, 2018

A puzzle posed by British puzzler Henry Ernest Dudeney. It was published in the Strand magazine in April 1930.

Smith, Jones and Robinson are the driver, fireman and guard on a train, but not necessarily in that order. The train carries three passengers, coincidentally with the same surnames, but identified with a “Mr.”: Mr. Jones, Mr. Smith and Mr. Robinson.

Mr. Robinson lives in Leeds.
The guard lives halfway between Leeds and Sheffield.
Mr. Jones’s salary is £1,000 2s. 1d. per annum.
Smith can beat the fireman at billiards.
The guard’s nearest neighbour (one of the passengers) earns exactly three times as much as the guard.
The guard’s namesake lives in Sheffield.

What is the name of the engine driver?

The salary amount of £1,000 2s. 1d., or one thousand pounds, two shillings and one penny, is significant only in that it is not evenly divisible by three.

[Answer to Puzzle #5 to follow in the next Puzzle]

Solution to Puzzle #4: Charlemagne’s Puzzle

Alcuin’s original solution involved seven steps:

  1. Take the sheep over
  2. Return – the sheep is on one side and the wolf and cabbage are on the other
  3. Take the cabbage over
  4. Return with the sheep – the cabbage is on one side and the sheep and wolf (and farmer) are on the other
  5. Take the wolf over
  6. Return – the wolf and cabbage are on one side and the sheep is on the other
  7. Take sheep over – all three have crossed over

Thus there are seven crossings, four forward and three back.

This river-crossing puzzle has spawned many “cosmetic” variations, such as fox, goose, and beans, and has appeared in the folklore of many lands. It appeared in the Simpsons episode Gone Maggie Gone with Homer Simpson trying to shuttle Maggie, Santa’s Little Helper, and a Jar of Rat Poison that Looked like Candy.

One of our more “waggish” commenters suggests a cosmetic variation of a selectman, a taxpayer, and the taxpayer’s money.

Puzzle #4: Charlemagne’s Puzzle

By Muriel Bristol (Transcriber) | September 18, 2018

The eighth-century English scholar Alcuin devised the following puzzle for the Emperor Charlemagne.

A traveler comes to a riverbank with a wolf, a goat and a head of cabbage. To his chagrin, he notes that there is only one boat for crossing over, which can carry no more than two passengers — the traveler and either one of the two animals or the cabbage. As the traveler knows, if left alone together, the goat will eat the cabbage and the wolf will eat the goat. The wolf does not eat cabbage. How does the traveler transport his animals and his cabbage to the other side intact in a minimum number of back-and-forth trips?

[Answer to Puzzle #4 to follow in the next Puzzle]

Solution to Puzzle #3: Lightbulbs in the Attic

Turn on a switch and leave it on for several minutes. Then turn it off and turn on a second switch. Go to the attic. One light is burning: the one that switched on second and is still active. Feel the two bulbs that are not burning. One of them is still warm from having been switched on by the first switch for several minutes. By a process of elimination, the remaining bulb (the cool one) is activated by the third switch that was never used.

Puzzle #3: Lightbulbs in the Attic

By Muriel Bristol (Transcriber) | September 11, 2018

One of many variants of the lightbulb puzzle:

There are three light switches downstairs. Each activates one of three lightbulbs in the attic. You can turn the switches on and off as much as you like and leave them in any position.

How can you tell which light switch corresponds to which lightbulb, if you are only allowed one trip upstairs?

[Answer to Puzzle #3 to follow in the next Puzzle]

Solution to Puzzle #2: Love in Kleptopia

Ms. Calderbrook’s own solution:

Jan sends Maria a box with the ring in it and one of his padlocks on it. Upon receipt Maria affixes her own padlock to box and mails it back with both padlocks on it. When Jan gets it he removes his padlock and sends the box back to Maria; voila! This solution is not just play; the idea is fundamental in Diffie-Hellman key exchange, an historic breakthrough in cryptography.

Other solutions are possible. One involved a second lock with a key small enough to fit inside the keyhole of a larger lock.

One of our commenters (Mike Sylvia) proposed a contemporary technological solution:

With the tip of the hat to Cody Wilson, I’ll take up a modern solution. Jan makes a scan of the key with which he padlocked the box. He e-mails the data file (encrypted, of course) to Maria. Maria uses her 3-D printer to reproduce the key to unlock the box when it arrives.

Yes, it seems like that would work too. It is not a perfect solution, in that it uses additional items, such as scanners and 3-D printers, not mentioned in the original puzzle. But, it is certainly novel and interesting.

Puzzle #2: Love in Kleptopia

By Muriel Bristol (Transcriber) | September 4, 2018

Caroline Calderbank, young daughter of mathematicians Ingrid Daubechies and Rob Calderbank, posed this problem.

Jan and Maria have fallen in love (via the internet) and Jan wishes to mail her a ring. Unfortunately, they live in the country of Kleptopia where anything sent through the mail will be stolen unless it is enclosed in a padlocked box. Jan and Maria each have plenty of padlocks, but none to which the other has a key. How can Jan get the ring safely into Maria’s hands?

[Answer to Puzzle #2 to follow in the next Puzzle]

Answer to Puzzle #1: Going to St. Ives

The traditional answer to Puzzle #1 is 1. Only the narrator is explicitly said to have been going to St. Ives.

All of the others, all 2,801 of them, are assumed traditionally to have been going the other way or standing along the way. (If one subtracts out the sacks, the living beings total 2,752).

Were they all headed to St. Ives, that would make 2,802 going there (including the narrator). (If you discount the sacks, the living beings total 2,753).