Black and White

kraemer chess problem

By Adolf Kraemer. White to mate in two moves.

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1. Rc8! Kxg3 2. Qc7# From Deutsche Tageszeitung, 1922.

September 25, 2015October 4, 2015 | Puzzles

Getting More Fiber

You’re locked on the roof of a building that’s 800 feet tall, and you want to reach the ground uninjured. You can safely fall only about 5 feet. The walls of the building are completely smooth and featureless, except that one wall has a small ledge that’s 400 feet above the ground. There’s a hook on this ledge, and another hook directly above it at the edge of the roof.

You have 600 feet of rope and a knife. How can you reach the ground?

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Cut the rope into lengths of 400 and 200 feet. You can do this by simply folding it into thirds or by measuring it against the building (tie one end to the top hook, climb down 400 feet to the ledge, cut the rope at that point, and climb back up). Now tie a loop in one end of the 200-foot rope. Pass the 400-foot rope through this loop so that half its length lies on either side. Tie the other end of the 200-foot rope to the top hook. The compound rope is 400 feet long, which is long enough to reach the ledge. Once you’re there, pull the 400-foot rope through the loop, tie it to the lower hook, and descend to the ground. That’s from Puzzling Stack Exchange. See “Stealing the Bell Ropes.”

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September 18, 2015September 27, 2015 | Puzzles

Heart and Head

Just offering this as a curiosity:

ohnishi cryptarithm

It was devised by the late Japanese cryptogramist Kyoko Ohnishi, and praised by puzzle maven Nob Yoshigahara. If each letter stands for a unique digit, what product is encoded here?

When this appeared in MIT Technology Review in November 2012, it attracted 20 responses, but it appears that all of them used computers to find the solution. If there’s a way to reason it out, I don’t think anyone has found it yet.

If we assume that neither of the factors has a leading zero, and that the partial products have five and four digits, as shown, then the solution is unique. I’ll put it in the spoiler box below, in case you want to work on it yourself.

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The intended solution seems to be 1572 × 39 = 61308. If we allow a four-digit number as the first partial product, then 3598 × 12 = 43176 also works. Yoshigahara himself passed away in 2004, so if this has a deductive solution, I’m not sure anyone knows what it is! The problem appeared in Allan Gottlieb’s Puzzle Corner in the November/December 2012 issue of the Review, the solution in the March/April 2013 issue.

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September 14, 2015September 27, 2015 | Puzzles

Black and White

techritz problem

By M. Techritz. Place the black and white kings on the board so that White can mate in one move.

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Place the black king on h1 and the white king on f3: Now 1. Kxf2 is mate.

September 12, 2015September 20, 2015 | Puzzles

In a Word

ditation
n. enrichment

improficuous
adj. unprofitable

Another puzzle by Yakov Perelman:

“Two fathers gave their two sons some money. One gave his son 150 rubles and the other 100 rubles. When the two sons counted their finances, they found that together they had become richer by only 150 rubles. What is the explanation?”

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“The whole trick is that one of the fathers is the son of the other father. There are only three persons in this problem and not 4: grandfather, father and son. Grandfather gave his son 150 rubles and the latter passed on 100 of them to the grandson (i.e. his son), thus increasing his own capital by 50 rubles.” From Figures for Fun, 1957.

September 9, 2015September 20, 2015 | Language · Puzzles

Dizzy

Russian science writer Yakov Perelman asks: How many times must an 8-toothed cogwheel rotate on its axis to circle around a 24-toothed cogwheel?

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Four. “Generally speaking, when a body rotates round a circle, it always does one revolution more than one can count. It is precisely this that explains why the earth, revolving round the sun, succeeds in rotating on its axis not in 365 1/4 days, but in 366 1/4 days, if one counts its revolutions in respect to the stars and not the sun. You will understand now why sidereal days are shorter than solar days.” From Perelman’s Figures for Fun, 1957.

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August 30, 2015September 6, 2015 | Puzzles · Science & Math

Black and White

tanner chess problem

By John Tanner. White to mate in two moves.

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1. f4! Kxe4 2. Rc4# From E.B. Cook et al., American Chess-Nuts, 1868.

August 29, 2015September 6, 2015 | Puzzles

Nobody Home

pooh and piglet

A puzzle by S. Sefibekov:

Winnie-the-Pooh and Piglet set out to visit one another. They leave their houses at the same time and walk along the same road. But Piglet is absorbed in counting the birds overhead, and Winnie-the-Pooh is composing a new “hum,” so they pass one another without noticing. One minute after the meeting, Winnie-the-Pooh is at Piglet’s house, and 4 minutes after the meeting Piglet is at Winnie-the-Pooh’s. How long has each of them walked?

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Call Winnie-the-Pooh’s house W and Piglet’s house P, and call their meeting point M. Suppose it took x minutes for the two of them to travel from their starting points to M. Then Winnie-the-Pooh spent x minutes walking from W to M and 1 minute from M to P; hence WM/MP = x. And the same reasoning for Piglet gives PM/MW = x/4. Now, WM/MP x PM/MW = 1, so 1 = x²/4, and x = 2. Winnie-the-Pooh has walked for 3 minutes and Piglet for 6 minutes. From the Russian science magazine Kvant.

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August 25, 2015September 6, 2015 | Puzzles

Bright and Early

https://commons.wikimedia.org/wiki/File:2014_Meteor.jpg — Image: Wikimedia Commons

Meteors are more commonly seen between midnight and dawn than between dusk and midnight. Why?

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After midnight, skywatchers occupy a part of the globe that faces “forward” along Earth’s trajectory around the sun, so the meteoroids they see enter the atmosphere at a relatively high velocity. Between dusk and midnight we are facing “backward,” so the meteors we observe are catching up with us and enter the atmosphere at a lower speed. The faster a meteor travels through the atmosphere, the brighter it appears.

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