Letters and Numbers

The first 10 letters of the alphabet, ABCDEFGHIJ, form a cipher that conceals the name of a number less than 100. What is the number?

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Eighty-four.

February 10, 2014February 16, 2014 | Puzzles

Groaners

How does a deaf man indicate to a hardware clerk that he wants to buy a saw?
How can you aim your car north on a straight road, drive for a hundred yards, and find yourself a hundred yards south of where you started?
What runs fore to aft on one side of a ship and aft to fore on the other?
A very fast train travels from City A to City B in an hour and a quarter. But the return trip, made under identical conditions, requires 75 minutes. Why?
Does Canada have a 4th of July?
Exhausted, you go to bed at 8 p.m., but you don’t want to miss an appointment at 10 a.m. the next day, so you set your alarm clock for 9. How many hours do you sleep?

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He says, “I want to buy a saw.” Drive backward. Its name. An hour and a quarter is 75 minutes. Yes. One.

February 9, 2014February 16, 2014 | Puzzles

Black and White

shinkman chess problem

By William Anthony Shinkman. White to mate in two moves.

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After the quiet 1. Qf2!, the queen will mate next move. From the Dubuque Chess Journal, July 1871.

February 7, 2014February 16, 2014 | Puzzles

The Rolling Die

rolling die puzzle

Imagine a die that exactly covers one square of a checkerboard. Place the die in the top left corner with the 6 uppermost. Now, by tipping the die over successively onto each new square, can you roll it through each of the board’s 64 squares once and arrive in the upper right, so that the 6 is exposed at the beginning and end but never elsewhere?

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The solution above is unique. I think this puzzle was devised by John Harris of Santa Barbara, Calif., in 1964; I found it in Miodrag Petkovic’s Mathematics and Chess, 1997.

February 5, 2014February 16, 2014 | Puzzles

Perspective

In a photograph, is there a way to distinguish between a landscape and its reflection?

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Yes. Because the camera is above the reflective surface, the two images are not identical. If the camera lens is, say, 5 feet above the water, then the reflected image appears as if viewed from 5 feet below it.

February 4, 2014February 9, 2014 | Puzzles

Black and White

holst chess problem

By Viktor Holst. White to mate in two moves.

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1. Qe4! and, one way or another, the queen will mate. From the Illustrated Family Journal, 1917.

January 31, 2014February 9, 2014 | Puzzles

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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