Dispatch

By Sam Loyd. In how few moves can White mate?

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White can mate in one move — bxa8=Q#:

September 24, 2009January 22, 2013 | Puzzles

Find the Theme, Part 1

What do these words have in common?

BEANED
DOTTED
GRANTED
HERBAL
HOMERED
JACKAL
LEEWARD
ROYAL
PATRON
VICTIM
VICTORIAN

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Each word can be broken into a pair of given names (BEA NED, etc.).

September 4, 2009January 22, 2013 | Puzzles

Spared

spared

By Karl Fabel. White to move and not checkmate.

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1. Rc6+ permits 1. … Rxh7. From Rätselstunde, 1952.

August 18, 2009September 11, 2015 | Puzzles

Royal Pain

You’re a knight in love with a princess. Unfortunately, the king knows you’re poor and disapproves of the match.

On the night of a great feast, the king calls you up before his men and presents a golden box. In it are two folded slips of paper. One, he announces, reads “Marriage,” the other “Death.” “Choose one,” he says.

Pretending to stir the fire, the princess manages to whisper that both slips say “Death.” But the king and his men are waiting, and you cannot escape now.

What should you do?

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Take either slip and drop it into the fire. Say, “Oops! I didn’t manage to read my fate. But we can soon determine what it was — what does the remaining slip say?”

August 5, 2009January 21, 2013 | Puzzles

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