In a Word

manzil
n. the distance between two stopping places

Another puzzle by Sam Loyd: Two ferry boats ply the same route between ports on opposite sides of a river. They set out simultaneously from opposite ports, but one is faster than the other, so they meet at a point 720 yards from the nearest shore. When each boat reaches its destination, it waits 10 minutes to change passengers, then begins its return trip. Now the boats meet at a point 400 yards from the other shore. How wide is the river?

“The problem shows how the average person, who follows the cut-and-dried rules of mathematics, will be puzzled by a simple problem that requires only a slight knowledge of elementary arithmetic. It can be explained to a child, yet I hazard the opinion that ninety-nine out of every hundred of our shrewdest businessmen would fail to solve it in a week. So much for learning mathematics by rule instead of common sense which teaches the reason why!”

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At their first meeting, 720 yards from one shore, the boats together have traveled a distance equal to the width of the river. When each reaches its destination, the combined distance is twice the width of the river. At their second meeting, 400 yards from the other shore, the combined distance is three times the river’s width. So at the second meeting each boat has gone three times as far as it had at the first meeting. At the first meeting one boat had traveled 720 yards, so at the second meeting it has gone three times this distance, or 2,160 yards. At this point it has covered 400 yards since doubling back, so the river is 2,160 – 400 = 1,760 yards, or one mile, wide. “The amount of time each ship consumed at the landing does not affect the problem.”

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January 30, 2014February 9, 2014 | Language · Puzzles

Back from the Klondike

Sam Loyd devised this puzzle in 1898. Begin at the heart in the center and move three squares in any of the eight directions, north, south, east, west, northeast, northwest, southeast, or southwest. You’ll land on a number; take this as the length of your next “march,” which again can go in any of the eight directions. “Continue on in this manner until you come upon a square with a number which will carry you just one step beyond the border, thus solving the puzzle.”

Interestingly, Loyd devised this puzzle expressly to defeat Leonhard Euler’s method of solving mazes. “Euler, the great mathematician, discovered a rule for solving all manner of maze puzzles, which, as all good puzzlists know, depends chiefly upon working backwards. This puzzle, however, was built purposely to defeat Euler’s rule and out of many attempts is probably the only one which thwarts his method.” The original puzzle, as published in the New York Journal and Advertiser, contained a flaw that permitted multiple solutions. That’s been corrected here — there’s only one way out.

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Move: 3 squares southwest to 4. 4 squares southwest to 6. 6 squares northeast to 6. 6 squares northeast to 2. 2 squares northeast to 5. 5 squares southwest to 4. 4 squares southwest to 4. 4 squares southwest to 4. 4 squares southeast “to liberty.” UPDATE: Wait, there’s at least one additional solution — in the last step you can go northwest as well as southeast. (Thanks, Carlos.) Now I wonder whether there’s ever been a version with a unique solution.

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January 27, 2014February 2, 2014 | Puzzles

Double Trouble

Two-move chess is just like regular chess except that each side makes two moves at a time. Prove that White, who goes first, can be sure of at least a draw.

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Well, suppose that isn’t the case. This means that Black has a strategy that’s guaranteed to win no matter what White plays. But imagine that, on his first pair of moves, White moves a knight out onto the board and then returns it to its home square. Now Black finds himself in White’s shoes: All the pieces are in their original positions and it’s his turn to move. Our hypothesis says that anyone in these circumstances is sure to lose, but it also says that Black is sure to win. This is a contradiction, so the hypothesis must be wrong: It can’t be the case that White is sure to lose. Hence he must be able to achieve at least a draw. From Svetoslav Savchev, Mathematical Miniatures, 2003.

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January 25, 2014February 2, 2014 | Puzzles

Black and White

carpenter chess problem

By George Edward Carpenter. White to mate in two moves.

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1.Rd4! Kg5 2.Qg6# From American Chess Magazine, 1897.

January 23, 2014February 2, 2014 | Puzzles

Black and White

drnek chess problem

By Josef Drnek. White to mate in two moves.

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1. Qe1! and the queen will mate. From Nové Parížské Mody, 1904.

January 17, 2014January 26, 2014 | Puzzles

An Epigrammatic Puzzle

Nokes went, he thought, to Styles’s wife to bed,
Nor knew his own was laid there in her stead;
Civilian, is the child then begot
To be allow’d legitimate or not?

— Bon Ton Magazine, July 1794

January 16, 2014January 15, 2014 | Puzzles

Thought That Counts

Chris Maslanka devised this brainteaser for the Gathering for Gardner held in Atlanta in April 2004:

A bouquet contains red roses, whites roses, and blue roses. The total number of red roses and white roses is 100; the total number of white roses and blue roses is 53; and the total number of blue roses and red roses is less than that.

How many roses of each color are there?

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Call the number of red roses, white roses, and blue roses r, w, and b, respectively. So r + w = 100; w + b = 53; and b + r = x, where x is less than 53. Adding these three equations gives 2r + 2b + 2w = 153 + x. The expression on the left is an even number, so 153 + x must be even as well. That means x is odd. Since 2w + 2r = 200, we know that 200 + 2b = x + 153 and thus x = 47 + 2b. We know that the bouquet contains blue roses (that is, b is not zero) and that x is less than 53, so x is an odd number higher than 47 and lower than 53. So it’s 49 or 51. If x = 49, then adding the three equations and dividing by 2 gives r + b + w = 101, and we have 1 blue rose, 52 white roses, and 48 red roses. But the problem states that the bouquet contains blue roses in the plural. So x must be 51, and we have 2 blue roses, 51 white roses, and 49 red roses.

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January 15, 2014January 26, 2014 | Puzzles

Black and White

meyer chess problem

By Heinrich Friedrich Ludwig Meyer. White to mate in two moves.

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1. Qh4! and the queen will mate from e1 or d8. From The Chess Amateur, April 1916.

January 10, 2014January 19, 2014 | Puzzles

Digit Jump

digit jump puzzle

Place the digits 1 through 8 into these circles so that no two successive numbers are connected by a line. If we don’t count rotations or reflections, the solution is unique.

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One way to solve this is to invert the task: Draw a new diagram in which each circle is connected only to the circles to which it had not previously been connected, and then seek a connected path along which the successive digits 1-8 can be placed. There are four of these, all rotations and reflections of the solution above.

January 9, 2014January 19, 2014 | Puzzles

Black and White

cumpe chess problem

By Josef Cumpe, 1908. White to mate in two moves.

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White wins by passing his turn with 1. Kd6! Now every Black move exposes some fatal vulnerability: If he moves the bishop then White will mate with queen or knight, and if he moves the knight then White can capture the bishop. Note that d6 is the only safe square for the king’s waiting move — every other option invites some troublesome mischief (1. Kc5 Bg1!).

January 3, 2014January 12, 2014 | Puzzles