Cross Purposes

Three men, A, B, and C, are given a test in quick thinking. Each man’s forehead is marked with either a blue or a white cross, and they’re put into an empty room. None of the three can see the color of his own cross, and they aren’t allowed to communicate in any way. Each is told that he can leave the room if he either sees two white crosses or can correctly deduce the color of his own cross.

The men know each other well, and A knows he’s just a bit more alert than the others. He sees that both B and C have blue crosses, and after a moment’s thought he’s able to leave the room, having correctly named the color of his own cross. What was the color, and how did he deduce it?

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His cross was blue. He thought, “If my cross were white, then B would see one blue and one white cross. In that event B would immediately know that his own cross could not be white, because if it were then C would see two white crosses and immediately leave the room. Similarly, if my cross were white then C would know that his own cross must be blue, since otherwise B would have gone out. Since neither B nor C has moved, my cross must be blue.”

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April 5, 2017April 16, 2017 | Puzzles

Black and White

de veer chess problem

By J.R.G. De Veer. White to mate in two moves.

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1. Qg3! Ke7 2. Qd6# From Jeugdproblem, 1914.

April 3, 2017April 16, 2017 | Puzzles

77777 …

A curious puzzle from the Penguin Problems Book, 1940:

A certain number consisting entirely of 7s is divisible by 199. Find the last four digits of the quotient, without finding the entire quotient.

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199 × the quotient equals the number we’re seeking (7777 …). Hence (200 × the quotient) minus the quotient will give us the same number. So let a, b, c, and d be the last four digits of the quotient. Now the quotient multiplied by 200 ends in r, s, 0, 0, where r, s are the digits that result from multiplying c, d by 2. Here’s the subtraction: So d must be 3, and c must be 2. Therefore s, which we know is twice d, is 6, and r, which is twice c, is 4. And that tells us that b is 8 and a is 6. So the last four digits of the quotient are 6823.

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April 2, 2017April 9, 2017 | Puzzles

Quickie

A puzzle by Frank Morgan: The meaning of a common English word becomes plural when an A is added at its start. What is the word?

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YES becomes AYES. From MIT Technology Review, July/August 2007.

March 29, 2017April 9, 2017 | Puzzles

A Hat Puzzle

khovanova hats

A puzzle by MIT mathematician Tanya Khovanova:

Three logicians walk into a bar. Each is wearing a hat that’s either red or blue. Each logician knows that the hats were drawn from a set of three red and two blue hats; she doesn’t know the color of her own hat but can see those of her companions.

The waiter asks, “Do you know the color of your own hat?”

The first logician answers, “I do not know.”

The second logician answers, “I do not know.”

The third logician answers, “Yes.”

What is the color of the third logician’s hat?

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If Logician #1 saw that both of her companions were wearing blue hats, then she’d know that her own hat must be red. But she says, “I do not know.” That tells us that either #2 or #3 (or both) is wearing a red hat. Logician #2 understands this. So if she saw that #3 was wearing a blue hat, then she’d know that she herself was wearing a red one. But she too says, “I do not know.” That means that Logician #3 must be wearing a red hat. Logician #3, realizing this, says so. Interestingly, Khovanova says her puzzle was inspired by a joke: Three logicians walk into a bar. The waitress asks, “Do you all want beer?” The first logician answers, “I do not know.” The second logician answers, “I do not know.” The third logician answers, “Yes.” She suggests that it might be valuable to find a way to convert jokes into puzzles and vice versa. (Tanya Khovanova, “Hat Puzzles,” Recreational Mathematics Magazine 1:2 [September 2014], 5-10.)

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March 28, 2017April 9, 2017 | Puzzles

A Mathless Math Puzzle

$hess bug puzzle$

Richard Hess posed this problem in the Spring 1980 issue of Pi Mu Epsilon Journal. At noon on Monday, a bug departs the upper left corner, X, of a p × q rectangle and crawls within the rectangle to the diagonally opposite corner, Y, arriving there at 6 p.m. He sleeps there until noon on Tuesday, when he sets out again for X, crawling along another path within the rectangle and reaching X at 6 p.m. Prove that at some time on Tuesday the bug was no farther than p from his location at the same time on Monday.

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Well, suppose there are two bugs making their journeys simultaneously — one departs X as the other departs Y, each headed for the opposite corner, and they travel different paths within the rectangle, arriving at their destinations simultaneously. At some time during their journeys they’ll both be the same horizontal distance from the left side of the rectangle. Since both paths remain within the rectangle, at that moment the distance between them can’t be greater than p. The editors received this answer from six readers, including Hess. “Their solutions were characterized by the complete absence of mathematical symbols and mathematical jargon,” they noted. “While there is certainly no objection to mathematical solutions to mathematical problems, a simple word-solution intelligible to any layman is to be preferred. Some of the other submitted solutions were profuse with subscripts, coordinates, inequalities, vinculi, functional relations, intermediate value theorems, Greek symbols, graphs, continuous functions and derivatives — all reminiscent of the sledgehammer method of swatting a fly.” The original problem is here (PDF, page 135).

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March 19, 2017July 17, 2017 | Puzzles

Pentalpha

During a visit to Crete in 1938, Miss L.S. Sutherland described a game she saw played on a pentagram:

You have nine pebbles, and the aim is to get each on one of the ten spots. You put your pebble on any unoccupied spot, saying ‘one’, and then move it through another, ‘two’, whether this spot is occupied or not, to a third, ‘three’, which must be unoccupied when you reach it; these three spots must be in a straight line. If you know the trick, you can do this one-two-three trick, for each of your nine pebbles and find it a berth, and then you win your money. If you don’t know the trick, it’s extremely hard to do it.

To make this a bit clearer: The figure has 10 “spots,” the five points of the star and the five corners of the pentagon in the middle. A move consists of putting a pebble on any unoccupied spot, moving it through an adjacent spot (which may be occupied) and continuing in a straight line to the next adjacent spot, which must be unoccupied. You then leave the pebble there and start again with a new pebble, choosing any unoccupied spot to begin this next move. If you can fill 9 of the 10 spots in this way then you’ve won.

Can you find a solution?

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Image: Wikimedia Commons One solution that’s easy to remember is to start your first move at a corner of the pentagon, arriving at a point of the star, and then contrive each subsequent move to end at the start of the preceding move (as above). Another solution:

March 17, 2017March 26, 2017 | Puzzles

Black and White

By T.R. Dawson. “The problem is for each White man to capture the corresponding Black man in such a way that the routes traversed by each man never come in contact with one another.”

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From Strand Magazine, October 1911.

March 14, 2017March 26, 2017 | Puzzles

Ships That Pass

For four months in 1840 Edgar Allan Poe conducted a puzzle column for the Philadelphia newspaper Alexander’s Daily Messenger. In that time he defied his readers to send him a cryptogram that he could not solve, and at the end of his tenure he declared himself undefeated. One of the later challenges came from 17-year-old Schuyler Colfax of New Carlisle, Iowa, who would grow up to become vice president of the United States:

Dear Sir — As you have in your Weekly Messenger defied the world to puzzle you by substituting arbitrary signs, figures, etc. for the different letters of the alphabet, I have resolved to try my utmost to corner you and your system together, and have manufactured the two odd looking subjects which accompany this as avant couriers. … If you succeed in solving the accompanying, I will, of course, as you request, acknowledge it publicly to my friends.

Poe responded: “We have only time, this week, to look at the first and longest cypher — the unriddling of which, however, will no doubt fully satisfy Mr. Colfax that we have not been playing possum with our readers.” Here’s Colfax’s cryptogram:

8n()58†d w!0 b† !x6n†z k65 !nz k65,8l†n b)x 8nd)Pxd !zw8x 6k n6 36w-†nd!x86n;

x=†0 z†,5!z† x=† w8nz 8n 8xd 62n †dx††w !nz k653† 8x x6 5†36l†5 8xd P†l†P b0 5†l†n,†.

()n8)d

What’s the solution?

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“Injuries may be atoned for and forgiven, but insults admit of no compensation; they degrade the mind in its own esteem, and force it to recover its level by revenge.” — Junius