Technicalities

In presenting the rules of chess, some writers carelessly say that a pawn that reaches the eighth rank can be promoted to any piece that the player chooses. That’s a bit too generous, as a couple of puzzle composers have noted. In 1941 Leonid Kubbel presented this problem — White is to mate in two moves:

kubbel promotion puzzle

It’s not immediately clear how to release Black from his stalemate and still mate him on the next move. The solution is to promote the e7 pawn to a black king!

kubbel promotion puzzle - solution

Now it’s Black’s move — he has to play 1. … Kd8, and White can mate both kings with 2. Qd7#!

The Polish master Johannes Zukertort offered this one: White is to mate on the move:

zukertort promotion puzzle

Here White promotes the pawn to a black knight, ending the game. (Note that it must be a knight — crazy as it seems, this is the only black piece that produces mate.)

June 10, 2016June 9, 2016 | Puzzles

Divide and Conquer

Facing dental surgery one day, mathematician Matt Parker asked Twitter for a math puzzle to distract him. A friend challenged him to put the digits 1-9 in order so that the first two digits formed a number that was a multiple of 2, the first three digits were a multiple of 3, and so on.

Leaving the digits in the conventional order 1234356789 doesn’t work: 12 is divisible by 2 and 123 by 3, but 1234 isn’t evenly divisible by 4. “By the end of my dental procedure, I had some but not all of the digits worked out, but, apparently, you’re not allowed to stay in the dentist’s chair after they’re finished.” At home he finished working out the solution, which is unique. What is it?

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381,654,729. (From Parker’s 2014 book Things to Make and Do in the Fourth Dimension.)

June 9, 2016June 19, 2016 | Puzzles

Not So Fast

José Paluzie offered this chess poser in 1910. White is to move and mate in 1:

paluzie chess puzzle

The key is to notice that the position is illegal: There’s no legal way for the black king to have arrived at a2. Black, desperate to avoid mate, must have put it there when White wasn’t looking.

Where did it come from? It doesn’t matter: White can place the black king on any legal square and mate in 1.

(From Burt Hochberg’s Chess Braintwisters, 1999.)

May 23, 2016May 23, 2016 | Puzzles

Balance

balance puzzle

Point P lies within acute angle XOY. How can we find a point A on OX and a point B on OY such that P is the midpoint of a segment drawn between them?

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Draw OP and continue the line to Q so that OP = PQ. Then draw QA parallel to OY and QB parallel to OX, as shown. OAQB is a parallelogram, so its diagonals bisect one another, and thus P is the midpoint of AB. From Paul Sloane and Des MacHale’s Mathematical Lateral Thinking Puzzles, 2015.

May 21, 2016May 29, 2016 | Puzzles

A Rolling Lemon

My lousy car has an odometer without 4s — in every position, the counter advances from 3 directly to 5. For example, when it read 000039 I drove one mile and watched it roll over to 000050. Today the odometer reads 002005. How many miles has the car actually traveled?

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Because it’s using 9 digits, the odometer is recording the mileage in base 9, except that its digits 5, 6, 7, 8, and 9 represent the base-9 digits 4, 5, 6, 7, and 8. So the actual (base 10) mileage today is just 2004₍₉₎, or 2 × 9³ + 4 = 2 × 729 + 4 = 1462. (From the 2005 American Mathematics Competition.)

May 12, 2016May 22, 2016 | Puzzles

Left or Right?

bicycle puzzle

You come upon the track of a bicycle in the mud. Was the bicycle traveling to the left or the right?

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On a conventional bicycle the rear wheel is fixed — at any given moment it’s traveling toward the front wheel’s point of contact with the ground. This means that a tangent drawn to the rear wheel’s track will always intersect the front wheel’s track. The reverse is not necessarily true — in the diagram above, the yellow tangent drawn to one curve doesn’t intersect the other curve at all. So the curve with the yellow tangent must correspond to the front wheel. Now, which way was the bicycle going? Choose some points on the rear wheel’s curve and draw tangents at those points, extending them in both directions until they reach the front wheel’s curve. This corresponds to putting the actual bicycle on the diagram, with its rear wheel at the tangent point, and assessing the position the front wheel would take in either direction, left and right. The distance between the wheels is constant, so we want to choose the direction in which the length of the segments is unchanging. Above, the leftward segments of the blue and green tangents are the same length, but the rightward segments are different, so the bicycle must have been traveling from right to left.

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May 10, 2016May 22, 2016 | Puzzles

Black and White

mikkelsen chess problem

By Paul Helweg Mikkelsen. White to mate in two.

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1. Be4! and the dark-squared bishop will mate. From Skakbladet, July 1912.

May 2, 2016May 15, 2016 | Puzzles

Pagan Island

Twenty-six villages are ranged around the coastline of an island. Their names, in order, are A, B, C, …, Z. At various times in its history, the island has been visited by 26 missionaries, who names are also A, B, C, …, Z. Each missionary landed first at the village that bore his name and began his work there. Each village was pagan to begin with but became converted when visited by a missionary. Whenever a missionary converted a village he would move along the coastline to the next village in the cycle A–B–C-…-Z–A. If a missionary arrived at an uncoverted village he’d convert it and continue along the cycle, but there was never more than one missionary in a village at a time. If a missionary arrived at a village that had already been converted, the villagers, feeling oppressed, would kill him and revert to a state of paganism; they would do this even to a missionary who had converted them himself and then traveled all the way around the island. There’s no restriction as to how many missionaries can be on the island at any given time. After all 26 missionaries have come and gone, how many villages remain converted?

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None. Each village is converted by its namesake missionary, unless it has already been converted by someone else, so every village gets converted at some point. But then each becomes a deathtrap for the next missionary who visits it. As each missionary works his way around the island, he leaves a string of such traps behind him, the first of which will spring on him when he comes around again unless he’s killed even sooner by stumbling on a village converted by someone else. So all 26 missionaries must die. Since a murderous village reverts to a pagan state, over time the 26 deaths provide 26 unconverted villages, but these are not necessarily different. If we could show that once a village has killed a missionary it is never visited again, and is thus effectively removed from the story, then the 26 deaths leave behind 26 different unconverted villages. So suppose village A has just murdered its first victim. That means it has been visited twice, once by the missionary who converted it from paganism and then by the missionary whom it murdered. What is the possibility that it will receive a third visit? Well, one of the two known visitors might be missionary A on his first visit to the island. But any other visitor to A must come from village Z. So if A were to get 3 visitors, then at least 2 of them must have come from Z; that is, at least 2 missionaries must have got into and out of village Z. But Z kills every other missionary who visits it. So if 2 missionaries left Z, then at least 3 must have entered it. Thus in order for A to receive a third visitor, Z must already have had at least 3 visitors. By the same reasoning, Y must have had at least 3 visitors, and so on around the island, until we reach the impossible condition that in order to have a third visitor, village A must already have had 3 visitors. This shows that A cannot receive a third visitor, and the conclusion follows. (From Crux Mathematicorum, via Ross Honsberger’s More Mathematical Morsels, 1991.)

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None. Each village is converted by its namesake missionary, unless it has already been converted by someone else, so every village gets converted at some point. But then each becomes a deathtrap for the next missionary who visits it. As each missionary works his way around the island, he leaves a string of such traps behind him, the first of which will spring on him when he comes around again unless he’s killed even sooner by stumbling on a village converted by someone else. So all 26 missionaries must die. Since a murderous village reverts to a pagan state, over time the 26 deaths provide 26 unconverted villages, but these are not necessarily different. If we could show that once a village has killed a missionary it is never visited again, and is thus effectively removed from the story, then the 26 deaths leave behind 26 different unconverted villages.

So suppose village A has just murdered its first victim. That means it has been visited twice, once by the missionary who converted it from paganism and then by the missionary whom it murdered. What is the possibility that it will receive a third visit? Well, one of the two known visitors might be missionary A on his first visit to the island. But any other visitor to A must come from village Z. So if A were to get 3 visitors, then at least 2 of them must have come from Z; that is, at least 2 missionaries must have got into and out of village Z. But Z kills every other missionary who visits it. So if 2 missionaries left Z, then at least 3 must have entered it. Thus in order for A to receive a third visitor, Z must already have had at least 3 visitors. By the same reasoning, Y must have had at least 3 visitors, and so on around the island, until we reach the impossible condition that in order to have a third visitor, village A must already have had 3 visitors. This shows that A cannot receive a third visitor, and the conclusion follows.

(From Crux Mathematicorum, via Ross Honsberger’s More Mathematical Morsels, 1991.)

April 24, 2016May 1, 2016 | Puzzles

Black and White

blumenthal chess problem

By Oskar Blumenthal. White to mate in two moves.

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1.Qh1! and White will mate with 2. Qh8# or 2. c7#. From Schweizerische Schachzeitung, 1907.

April 12, 2016April 24, 2016 | Puzzles

The Barbershop Paradox

In 1894 Lewis Carroll published a conundrum that, he wrote, presents “a very real difficulty in the Theory of Hypotheticals.” Suppose that Allen, Brown, and Carr run a shop. At least one of them must always be present to mind the shop, and whenever Allen leaves he always takes Brown with him. Now, suppose that Carr is out. In that case then if Allen is out then Brown must be in, in order to tend the shop. But we know that this isn’t true — we’ve been told that whenever Allen is out then Brown is out.

Since the supposition that Carr is out leads to a falsehood, then it must itself be false. Confusingly, the laws of logic seem to require that Carr never leave the shop.

“I greatly hope that some of the readers of Mind who take an interest in logic will assist in clearing up these curious difficulties,” Carroll wrote. Modern logicians would say that this is a simple error in reasoning, rather than a logical disaster. But what is the error?

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Consider the troublesome statement “If Allen is out, then Brown is in” (supposing that Carr is out). We had evaluated this as false. But saying “If Allen is out, then Brown is in” is just the same as saying “Either Allen is in or Brown is in.” And this is true, not false — it admits the possibility that Allen remains in the shop, as of course he must. So the problem disappears. In simple terms, when Carr leaves the shop then Allen (and perhaps Brown) stays behind, as our common sense says he does. (Lewis Carroll, “A Logical Paradox,” Mind 3:11 [July 1894], 436-438.)

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April 7, 2016April 17, 2016 | Puzzles