Black and White

keeble chess problem

By J. Keeble. White to mate in two moves.

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1. Ng5!, and the other knight will discover mate. From O.A. Brownson Jr., Chess Problems, 1876.

January 28, 2015February 8, 2015 | Puzzles

Threewise

http://commons.wikimedia.org/wiki/File:Napoleon's_theorem.svg — Image: Wikimedia Commons

Draw an arbitrary triangle and build an equilateral triangle on each of its sides, as shown.

Now show that AP = BQ = CR.

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Image: Wikimedia Commons Rotating the figure 60° counterclockwise around B carries A to R and P to C. And rotating it 60° clockwise around A carries B to R and Q to C. So AP = BQ = CR. From Edward Barbeau, Murray Klamkin, and William Moser, Five Hundred Mathematical Challenges, 1995.

January 23, 2015February 1, 2015 | Puzzles

Black and White

By Theophilus A. Thompson. White to mate in two moves.

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1. Ra4+! Bxa4 2. b4# From Theophilus A. Thompson, Chess Problems, 1873.

January 19, 2015January 25, 2015 | Puzzles

Ice Work

Three hockey pucks, A, B, and C, lie in a plane. You make a move by hitting one puck so that it passes between the other two in a straight line. Is it possible to return all the pucks to their original positions with 1001 moves?

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No. When viewed from above the three pucks form triangle ABC. With each hit the orientation of the triangle changes, so that sometimes “ABC” reads clockwise, sometimes counterclockwise. The original position was reached after zero hits, so it can be restored only after an even number of hits. From Arthur Engel, Problem-Solving Strategies, 1998.

January 15, 2015January 25, 2015 | Puzzles

Knights and Scoundrels

A problem from the 1994 Italian Mathematical Olympiad:

Every inhabitant on the island of knights and scoundrels is either a knight (who always tells the truth) or a scoundrel (who always lies). A visiting journalist interviews each inhabitant exactly once and gets the following answers:

A₁: On this island there is at least one scoundrel.
A₂: On this island there are at least two scoundrels.
…
A_n-1: On this island there are at least n – 1 scoundrels.
A_n: On this island everyone is a scoundrel.

Can the journalist decide whether the knights outnumber the scoundrels?

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I don’t have the olympiad solution, but it seems to me that there must be an equal number of knights and scoundrels. The first half of the interviewees are knights, and the second half are scoundrels. If there were more than n/2 knights, then they’d crowd out the scoundrels and their greater claims could not be true. If there were more than n/2 scoundrels, then they’d crowd out the knights and their lesser claims could not be false. UPDATE: Also, n must be even. If it’s odd, then the middle interviewee can be neither a knight nor a scoundrel, which is impossible. Thanks to those who wrote in to point this out.

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January 10, 2015January 18, 2015 | Puzzles

“A Bit of Spanish”

2015-01-07-a-bit-of-spanish

This pleasing cryptarithm, by Bob High, appears in the September/October 2014 issue of MIT Technology Review. If each letter stands for a digit, what arithmetic sum is enciphered here?

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POCO is 4595 and MUCHO is 68925. There are 15 POCOs, so the sum of each column is a multiple of 15, and hence (without the carry from the column to its right) ends in a 5 or a 0. The rightmost column has no carry, so the O in MUCHO stands for 0 or 5. Which is it? If it’s 0, that means that the sum of all the Os in the rightmost column is 0, and that there’s no carry into the tens column. If there’s no carry, we need to find a digit for C that when multiplied by 15 produces a two-digit answer (the CH in MUCHO) that starts with that digit (since the hundreds column in the sum is otherwise zero). The only digit that will do that is 1 (1 × 15 = 15). But if CH is 15 then there’s no carry into the thousands column, which means that U must be 0 or 5 (15 × P). And we’ve already assigned 0 to O and 5 to H. So this can’t be right. If it’s 5, then the carry into the tens column is 7 (15 × 5 = 75). What is C? Every possibility from 0 to 8 produces conflicts or requires C to have two different values, but C = 9 produces a sum in the tens column of 142 (15 × 9 plus the 7 carry). In that case H = 2 and 14 is carried into the hundreds column. We know that O is 5, so that column now totals 89 (5 × 15 + 14). C is 9 and 8 is carried into the thousands column, where it’s added to 15 Ps. Here every possibility except P = 4 creates duplications or conflicts, so U is 8 and M is 6. “Because O must equal 5 and C must equal 9 and P must equal 4, I believe the solution is unique,” writes NYU computer scientist Allan Gottlieb, who conducts the puzzle column. “I like these alphabetical number problems; but, cannot recall seeing this one before.”

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January 7, 2015January 18, 2015 | Puzzles

Black and White

agnel chess problem

By H.R. Agnel. White to mate in two moves.

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1. Bc8! Kg4 2. Re4# From Benjamin Glover Laws, The Two-Move Chess Problem, 1890.

January 6, 2015January 11, 2015 | Puzzles

Progress

The Martian parliament consists of a single house. Every member has three enemies at most among the other members. Show that it’s possible to divide the parliament into two houses so that every member has one enemy at most in his house.

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Consider all the possible divisions, and in each case count up the total number of cases E in which a member has an enemy in his own house. The division in which E is minimized must have the required property. If it didn’t, that would mean that some member has at least two enemies in his own house. In that case he’d have one enemy at most in the other house, and could reduce E further by moving to the opposite house. That’s a contradiction, so the required task must be possible.

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January 2, 2015January 11, 2015 | Puzzles

Red and Black

Take two decks of cards, minus the jokers, shuffle them together, and divide them into two piles of 52 cards. What is the probability that the number of red cards in the Pile A equals the number of black cards in Pile B? How many cards would you have to view to be certain of your answer?

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The probability is 100 percent. First suppose that Pile A is all red and Pile B all black. This fulfills the condition. Now move one red card from A to B. In order to keep 52 cards in each pile, we must also move a black card from B to A, so this new arrangement also fulfills the condition. And so on: So long as there are 52 cards of each color and 52 cards in each pile, the number of red cards in one pile must equal the number of black cards in the other. So you don’t have to view any cards at all. See Alcohol Problem.

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