Odd at Heart

A problem proposed by Charles W. Trigg in the Spring 1970 issue of the Pi Mu Epsilon Journal (Volume 5, Number 4):

The digits 1-9 can be arranged in a square array so that the digits in no column, row, or long diagonal appear in order of magnitude:

trigg pmej paritypuzzle

Prove that, however this is done, the central digit must be odd.

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One row, one column, and two diagonals pass through the center. In order to honor the requirement, we must arrange the eight outer digits so that none of these four triplets falls in ascending or descending order; that is, in each of these four triplets the center digit must be either the highest or the lowest value. In order to accomplish this, we must complete each triplet by adding two digits that are either both lower than the central digit or both higher. But if the center digit is even, that leaves us with an odd number of digits whose value is lower than the center digit and an odd number whose value is higher. So we’ll be forced to complete at least one of the four triplets by combining one higher and one lower value — producing a trio that falls in order of magnitude.

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December 19, 2018December 19, 2018 | Puzzles

Black and White

From Francis Healey, A Collection of Two Hundred Chess Problems, 1866. White to mate in two moves.

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White wins with the seemingly crazy 1. Qd4!: Black can’t prevent both 2. Rb6# and 2. Bf3#.

December 18, 2018 | Puzzles

Point Pairs

University College London mathematician Matthew Scroggs has released this year’s puzzle Christmas card for Chalkdust magazine.

This year’s card contains 10 puzzles, each with a numerical answer. Splitting each answer into two strings of digits will identify two dots on the card’s cover to be connected by a straight line. Drawing 10 lines correctly will produce a Christmas-themed picture.

You can download the card as a PDF or complete it online.

December 9, 2018 | Puzzles

The Indecisive Rower

A rower rows regularly on a river, from A to B and back. He’s got into the habit of rowing harder when going upstream, so that he goes twice as fast relative to the water as when rowing downstream. One day as he’s rowing upstream he passes a floating bottle. He ignores it at first but then gradually grows curious about its contents. After 20 minutes of arguing with himself he stops rowing and drifts for 15 minutes. Then he sets out after the bottle. After some time rowing downstream he changes his mind, turns around, and makes his way upstream again. But his curiosity takes hold once more, and after 10 minutes of rowing upstream he turns and goes after the bottle again. Again he grows ashamed of his childishness and turns around. But after rowing upstream for 5 minutes he can’t stand it any longer, rows downstream, and picks up the bottle 1 kilometer from the point where he’d passed it. How fast is the current?

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Between passing the bottle and picking it up, the rower spends a total of 20 + 10 + 5 = 35 minutes rowing upstream. In this period his total motion relative to the bottle (and to the water) is zero — the distance he’s traveled relative to the water’s surface is the same upstream as down. We know he moves half as fast relative to the water when he’s rowing downstream as when he’s rowing upstream; this means he’s spent twice as much time rowing downstream as up, or 70 minutes. He also spent 15 minutes drifting, so altogether he picks up the bottle 35 + 70 + 15 = 120 minutes, or two hours, after passing it. The bottle has covered 1 kilometer in this time, so the current is moving at 1/2 kilometer per hour. (A simpler way to think about the timing is to imagine that the whole thing takes place on a lake rather than a river. The rower departs from the bottle, rows 35 minutes in one direction, sits motionless for 15 minutes, and rows back to the bottle at half his earlier speed. When he reaches the bottle his round trip has taken him 35 + 15 + 70 = 120 minutes, or two hours.)

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December 5, 2018December 4, 2018 | Puzzles

Footwork

Each day after work, Smith takes the train to the suburb where he lives, and his chauffeur meets him at the station and drives him home. One day Smith finishes work early and arrives at the suburb one hour earlier than usual. He starts walking home, and the chauffeur meets him on the road and drives him the rest of the way. This gets Smith home 10 minutes earlier than usual. How long did he walk? (Disregard the time spent in stopping and picking up Smith, and assume that the chauffeur normally arrives at the station just as the train does.)

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The chauffeur has been spared the journey from the meeting point to the station and back. If this brings him home 10 minutes early, then it must normally take him 5 minutes to get from the meeting point to the station. Smith arrived at the station an hour early and arrived home 10 minutes earlier than usual, so it took him 50 minutes longer than usual to get home from the station. That’s due to his walking rather than riding from the station to the meeting point. This shows that it takes Smith 50 minutes more than the car to cover that distance. So he must have walked for 5 + 50 = 55 minutes. (From Fred Schuh, The Master Book of Mathematical Recreations, 1968.)

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November 28, 2018November 27, 2018 | Puzzles

Black and White

woodard chess problem

By Eugene Woodard. White to mate in two moves.

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1. Qa1! and the queen will mate. From the Dubuque Chess Journal, 1888.

November 25, 2018 | Puzzles

Illumination

In the Japanese logic puzzle Akari, you’re presented with a grid of black and white squares. The goal is to place “light bulbs” into white cells until the whole grid is illuminated. Each bulb sends out rays of light horizontally and vertically, illuminating its row and column unless a black cell blocks the rays.

There are two constraints: The bulbs must not shine on one another, and each numbered black cell must bear that many bulbs (orthogonally adjacent to it) in the finished diagram. An unnumbered black cell can bear any number of bulbs.

Here’s a moderately difficult puzzle. Can you solve it?

https://commons.wikimedia.org/wiki/File:Lightup.svg — Image: Wikimedia Commons

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Image: Wikimedia Commons

November 21, 2018November 21, 2018 | Puzzles

Ones and Twos

https://www.flickr.com/photos/evoo73/3222220257 — Image: Flickr

From the Kanja Otogi Zoshi (“Collection of Interesting Results”) of Nakane Genjun, 1743:

Your friend has 30 go stones. He lines them up out of your sight, placing down either one or two stones with each deposit and calling “here” so you’ll know this has been done. When all 30 stones have been placed, you have heard him say “here” 18 times. How many deposits contained one stone, and how many two?

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If the number of deposits of one stone is x, then the number of deposits of two stones is 18 – x. The total number of stones placed is x × 1 + (18 – x) × 2 = 30. Solving this gives x = 6, so in this case 6 deposits contained one stone and 12 contained two. More generally, if a is the total number of stones and n is the number of cries of “here,” then x × 1 + (n – x) × 2 = a x + 2n – 2x = a x = 2n – a Hence the number of times that two stones are placed is n – x = n – (2n – a) = a – n The number of deposits of one stone: 2n – a The number of deposits of two stones: a – n (From Sakurai Susumu, Wasan: The Fascination of Traditional Japanese Mathematics, 2009.)

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November 17, 2018 | Puzzles

Truth Pays

Here are a penny and a quarter. Make a statement. If your statement is true, then I’ll give you one of these coins (not saying which). But if your statement is false, then I won’t give you either coin.

Raymond Smullyan says, “There is a statement you can make such that I would have no choice but to give you the quarter (assuming I keep my word).” What statement will accomplish that?

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One statement that works is “You will not give me the penny.” If that were false, then I would give you the penny. But I’d said that if your statement were false I’d keep both coins. So your statement can’t be false; it must be true. And if you’ve made a true statement then I owe you a coin. So I have no choice but to give you the quarter. Smullyan related this puzzle to Gödel’s incompleteness theorem. “I thought of the penny as standing for provability and the quarter as standing for truth. Thus the statement ‘You will not give me the penny’ corresponds to Gödel’s sentence, which in effect says: ‘I am not provable.'” (From Logical Labyrinths, 2008.)

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November 12, 2018 | Puzzles

Winners and Losers

A puzzle by the Hungarian-Canadian mathematician George Grätzer:

I’m writing an article about a round-robin tennis tournament, in which each player plays each other player once. I decide to pick one player and ask her which players she defeated (in tennis there are no ties). Then I’ll ask each of those players which players they’ve defeated. Is it possible to pick a player so that everyone in the tournament is mentioned in my article?

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Yes. The simplest solution is to choose the tournament’s winner, A. Because she won, we know that no other player won more matches than she did. Now suppose I finish all my interviews and find there’s some player B I haven’t spoken with. By the rules of my own plan, that will mean that B lost neither to A nor to any of the players whom A defeated. But because there are no ties, that would mean that B won all those matches. And that would give B a higher total score than the tournament winner, a contradiction. So A must fulfill the condition. (From Train Your Brain, 2011.)

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November 11, 2018 | Puzzles