Rock and Roll

Worshipful natives are rolling a giant statue of me across their island. The statue rests on a slab, which rests on rollers that have a circumference of 1 meter each. How far forward will the slab have moved when the rollers have made 1 revolution?

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The slab will have moved forward 2 meters, not 1. If the slab were removed, 1 turn would advance the rollers 1 meter. If the rollers were held in place, 1 turn would advance the slab 1 meter. Combining the two motions means that 1 turn advances the slab 2 meters. See “The Famous ‘Wheel Question’.”

August 1, 2009January 21, 2013 | Puzzles

“Christopher Crusty’s Little Idea”

White to mate on the move.

“A real-right-down regular rare one. The problem exhibited is quite correct Chess, and no violation of any law takes place. In fact, it is found to be quite easy — when you know how.”

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The diagram is sideways. Bwahahaha! (ducks) How should we correct it? Well, if we turn it a quarter turn clockwise, then Black’s light-square bishop is trapped behind an unmoved pawn, which is impossible. So turn it to the left: Now, the black pawn on g2 must have started on h7; otherwise Black’s pawns must have devoured an impossible amount of material to reach their present positions. That means that the f5, f4, and g2 pawns cannot have moved last, and that means that Black’s last move must have been b7-b5. Thus White can capture the b-pawn en passant, giving mate: From the Leeds Mercury Supplement, collected in Thomas B. Rowland, Chess Fruits, 1884.

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July 25, 2009January 21, 2013 | Puzzles

Two by Two

Here’s a curious way to multiply two numbers. Suppose we want to multiply 97 by 23. Write each at the head of a column. Now halve the first number successively, discarding remainders, until you reach 1, and double the second number correspondingly in its own column:

two by two - first image

Cross out each row that has an even number in the left column, and add the numbers that remain in the second column:

two by two - second image

That gives the right answer (97 × 23 = 2231). Why does it work?

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Essentially the technique converts the first factor into binary, multiplies each of its constituents by the second factor, and sums the results. Imagine that each line is associated with a power of 2: the first line with 2⁰, the second with 2¹,and so on. The business in the first column, halving the first factor successively and crossing out those lines with even numbers, effectively reduces the first factor to its binary constituents — here, the lines that remain are those associated with 2⁰, 2⁵, and 2⁶, and, sure enough, 2⁰ + 2⁵ + 2⁶ = 97. Now we need to multiply each of those constituents by the second factor, 23. In other words, we want to find: (23 × 2⁰) + (23 × 2⁵) + (23 × 2⁶) That’s what’s accomplished in the second column. Doubling the second factor successively, as we’ve done there, is equivalent to multiplying it by 2⁰ in the first line, by 2¹ in the second, and so on. That is, 23 = 23 × 2⁰, 46 = 23 × 2¹, etc. And the lines that haven’t been crossed out are precisely the ones we want: 23 (= 23 × 2⁰) 736 (= 23 × 2⁵) 1472 (= 23 × 2⁶) So if we add those values, we’ll get the product of the original two numbers, which is what we sought: 23 + 736 + 1472 = 2231. Here’s essentially what we’ve done, from the top: 97 × 23 = (2⁰ + 2⁵ + 2⁶) × 23 = (23 × 2⁰) + (23 × 2⁵) + (23 × 2⁶) = 23 + 736 + 1472 = 2231 It works with any pair of numbers.

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July 18, 2009January 21, 2013 | Puzzles · Science & Math

“Paradox, by a Lady”

One summer evening, as I was walking in the fields, I heard somebody behind me calling out my name. I turned round, and saw a friend of mine, at the distance of 400 yards, approaching to join me. We each of us moved 200 yards, with our faces towards the other, in a direct line yet we were still 400 yards asunder. How could this be?

— The Nic-Nac; or, Oracle of Knowledge, Sept. 13, 1823

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“The lady walked 200 yards backward.”

July 16, 2009January 21, 2013 | Puzzles

Trade Relations

Northland and Southland were happy neighbors until yesterday, when Northland declared that a Southland dollar was to be worth only 90 Northland cents.

Not to be outdone, Southland declared that a Northland dollar would be worth 90 Southland cents.

I live in Centerville, on the border between the two countries. I go into a Northland store and buy a kazoo, which costs 10 cents. I pay for it with a Northland dollar and receive a Southland dollar as change.

Then I go across the street and enter a Southland store. There I buy a lemon, which also costs 10 cents. I pay for it with a Southland dollar and receive a Northland dollar as change.

When I get home I have my kazoo and lemon, for which it appears I’ve paid nothing. And each of the merchants has an additional 10 cents in his receipts.

So who paid for the kazoo and the lemon?

(From Eugene Northrop.)

July 11, 2009November 4, 2011 | Puzzles

“An Ingenious Match Puzzle”

From Henry Dudeney:

dudeney match puzzle

“Place six matches as shown, and then shift one match without touching the others so that the new arrangement shall represent an arithmetical fraction equal to 1. The match forming the horizontal fraction bar must not be the one moved.”

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“It will be seen that the second I in VII has been moved, so as to form the sign of square root. The square root of 1, is, of course, 1, so that the fractional expression itself represents 1.”

July 9, 2009January 21, 2013 | Puzzles

Figure and Ground

An island is a body of land surrounded by water, and a lake is a body of water surrounded by land.

Now suppose the northern hemisphere were all land, and the southern hemisphere water. Is one an island, or is the other a lake?

June 21, 2009 | Puzzles

The Pup Tent Problem

In 1980 the Educational Testing Service offered this question on an aptitude test:

In pyramids ABCD and EFGHI shown above, all faces except base FGHI are equilateral triangles of equal size. If face ABC were placed on face EFG so that the vertices of the triangles coincide, how many exposed faces would the resulting solid have?

(A) Five (B) Six (C) Seven (D) Eight (E) Nine

Which is correct?

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The test makers were expecting the answer (C), but several students pointed out that in fact (A) is correct — the shapes fit together to produce a surprisingly simple object with only five faces. This is easiest to see if you imagine two square pyramids side by side: The tetrahedron fits precisely between them to make a smooth-sided “pup tent”: If you have polyhedral dice, try fitting the face of a 4-sided to an 8-sided die; you’ll find that the adjoining sides are coplanar.

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June 12, 2009January 21, 2013 | Puzzles

A Curious Conversation

You’re standing with your friends Val and Colin when a stranger approaches and shows you 16 cards:

A♥ Q♥ 4♥
J♠ 8♠ 7♠ 4♠ 3♠ 2♠
K♣ Q♣ 6♣ 5♣ 4♣
A♦ 5♦

He shuffles the cards, selects one, and tells Val the card’s value and Colin the card’s color. Then he asks, “Do you know which card I have?”

Val says, “I don’t know what the card is.”

Colin says, “I knew that you didn’t know.”

Val says, “I know the card now.”

Colin says, “I know it too.”

What is the card?

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It’s the ace of diamonds. If Val couldn’t name the card knowing only its value, then it must be an ace, a queen, a 5, or a 4, as these are the only values that appear more than once. If Colin knew that Val couldn’t name it, it must be red, as all the red cards have values (A, Q, 5, 4) that appear more than once. When Val understands that the card must be red, she can exclude from her list of possibilities the queen, the 5, and the 4, as each of those cards appears in a black suit. That leaves only the ace of diamonds. Adapted from Julian Havil’s Impossible?, 2008. UPDATE: Wait, this is cooked. Havil gives the problem but not the solution, and my solution above isn’t valid. Thanks, Ben and Andrew, for pointing it out. If anyone can find a valid solution, let me know and I’ll post it here. I don’t think one is possible. FWIW, this is based on a more famous “impossible puzzle,” which is soluble but spectacularly hard. UPDATE 2: In Canada or England, does the word “colour” mean “suit”? If so, the problem is soluble. It seems to have appeared originally on page 14 of a paper [PDF] by University of Victoria philosopher Audrey Yap. The first two steps above are right — Val can deduce that the card is red and must be A, Q, 5, or 4. The fact that she’s able to say “I know the card now” at this point means it can’t be an ace (because that value appears in both red suits). And the fact that Colin can then say “I know it too” means that it must be the five of diamonds. So, from the top: Val says, “I don’t know what the card is.” (She knows it’s a 5, but there are two 5s.) Colin says, “I knew that you didn’t know.” (He knows it’s a diamond, and thus an A or 5; but as each of these values appears more than once in the deck, he knows that Val won’t have enough information yet to identify the suit.) Val says, “I know the card now.” (Colin’s remark has just told her that the card is red, because all the red cards in this deck have values that appear in multiple suits. And the only red 5 is 5♦.) Colin says, “I know it too.” (He knows that Val was looking for an A or a 5, and that learning the card’s color was helpful. Thus she can’t have been looking for an A, as both aces are red. That leaves 5♦.) So, if “colour” means “suit,” the answer is 5♦. If “colour” means “color,” then I still think it’s insoluble. Thanks to the many smart people who wrote in about this, especially Helge.

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It’s the ace of diamonds.

If Val couldn’t name the card knowing only its value, then it must be an ace, a queen, a 5, or a 4, as these are the only values that appear more than once.

If Colin knew that Val couldn’t name it, it must be red, as all the red cards have values (A, Q, 5, 4) that appear more than once.

When Val understands that the card must be red, she can exclude from her list of possibilities the queen, the 5, and the 4, as each of those cards appears in a black suit.

That leaves only the ace of diamonds.

Adapted from Julian Havil’s Impossible?, 2008.

UPDATE: Wait, this is cooked. Havil gives the problem but not the solution, and my solution above isn’t valid. Thanks, Ben and Andrew, for pointing it out. If anyone can find a valid solution, let me know and I’ll post it here. I don’t think one is possible.

FWIW, this is based on a more famous “impossible puzzle,” which is soluble but spectacularly hard.

UPDATE 2: In Canada or England, does the word “colour” mean “suit”? If so, the problem is soluble. It seems to have appeared originally on page 14 of a paper [PDF] by University of Victoria philosopher Audrey Yap. The first two steps above are right — Val can deduce that the card is red and must be A, Q, 5, or 4. The fact that she’s able to say “I know the card now” at this point means it can’t be an ace (because that value appears in both red suits). And the fact that Colin can then say “I know it too” means that it must be the five of diamonds. So, from the top:

Val says, “I don’t know what the card is.” (She knows it’s a 5, but there are two 5s.)

Colin says, “I knew that you didn’t know.” (He knows it’s a diamond, and thus an A or 5; but as each of these values appears more than once in the deck, he knows that Val won’t have enough information yet to identify the suit.)

Val says, “I know the card now.” (Colin’s remark has just told her that the card is red, because all the red cards in this deck have values that appear in multiple suits. And the only red 5 is 5♦.)

Colin says, “I know it too.” (He knows that Val was looking for an A or a 5, and that learning the card’s color was helpful. Thus she can’t have been looking for an A, as both aces are red. That leaves 5♦.)

So, if “colour” means “suit,” the answer is 5♦. If “colour” means “color,” then I still think it’s insoluble. Thanks to the many smart people who wrote in about this, especially Helge.

June 3, 2009September 24, 2017 | Puzzles

Don’t Even Try

White or Black to play and mate or self-mate in one move. That is, you must find a total of four moves from this position: a White move that mates Black instantly, a White move that forces Black to mate White instantly, and equivalent moves for Black.

“Memo: The above puzzle depends on a literal interpretation of the rule which provides that a Pawn on reaching the eighth square may become any piece irrespective of colour.”

WARNING: “This monstrosity is the production of an erratic solver who has been sorely tried, puzzled and perplexed all the year round by the many posers and problems which have appeared from time to time in the numerous Chess columns. His aesthetic patience, resignation, fortitude, culture and hope all at once breaking down, he set to work and with wrathful spirit, regardless of all problem construction, devised it more for the sake of retaliation and revenge than to give pleasure. To prove his spiteful character; when composing it, he was overheard repeating, ‘Since I cannot prove a lover to entertain these fair spoken days, I am determined to prove a villain.’ Consequently, gentle reader, we warn you not to attempt it, except indeed that you are the happy possessor of that knowledge wherein you are able to puzzle others. It may look beastly simple, but to any young solver who may be foolhardy enough to venture it we offer a few words of advice–carefully study the above memo and note that–but ‘hold enough,’ no more can we divulge, fearful of bringing the fiery wrath of the exasperated composer upon our devoted heads.”

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White: To mate, 1. Rb3#. To self-mate, 1. bxc8, promoting to a black king, pinning the pieces on b8 and c7 and leaving Black with only 1. … Kxc3#. Black: To mate, 1. … Kxc3#. To self-mate, 1. … Nxd5+, forcing 2. Nxd5#. “Notwithstanding the memo, many young solvers (and old ones too) have had a hard wrestle with this production, but all to no purpose, it was the first of its kind published. The consequence was that many communications were received by the composer doubting his sanity, denouncing the problem as a fraud and threatening that if they caught him on the other side of the channel they would give him a taste of their wrath. The receiver of these communications thought that the channel was a mighty fine affair!” From Thomas B. Rowland, Chess Fruits, 1884.

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May 28, 2009January 21, 2013 | Puzzles