Black and White

ropet chess problem

By Ivan Pavlovich Ropet. White to mate in two moves.

	Select Click for Answer
1. Nf4! (threatening 2. Rd5#) Kxd4 2.Bf6# From the St. Petersburg New Times, 1899.

August 16, 2015August 30, 2015 | Puzzles

Podcast Episode 69: Lateral Thinking Puzzles

Here are four new lateral thinking puzzles to test your wits! Solve along with us as we explore some strange situations using only yes-or-no questions.

See full show notes …

August 10, 2015October 24, 2017 | Podcast · Puzzles

Black and White

legentil chess problem

By Georges Legentil. White to mate in two moves.

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1. Nc4! stops the pawns, and the queen will mate. From Journal de Rouen, March 4, 1911.

August 3, 2015August 16, 2015 | Puzzles

Knife Act

knife act - problem

I have just baked a rectangular cake when my wife comes home and barbarically cuts out a piece for herself. The piece she cuts is rectangular, but it’s not in any convenient proportion to the rest of the cake, and its sides aren’t even parallel to the cake’s sides. I want to divide the remaining cake into two equal-sized halves with a single straight cut. How can I do it?

	Select Click for Answer
A line drawn through the center of a rectangle, in any direction, will cut it evenly in two. So a line that passes through both the center of the cake and the center of the removed rectangle will divide the remaining cake into equal halves, regardless of the proportions involved or the hole’s orientation. You can find the center of each rectangle by noting the intersection of its diagonals.

July 29, 2015August 9, 2015 | Puzzles

Black and White

ischty chess problem

By Juri Ischty. White to mate in two moves.

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1. Qh3! stops the h-pawn, and the queen will mate next move. From Shakhmatnoe Obozrenie, 1893.

July 22, 2015August 2, 2015 | Puzzles

Close Dosage

You’re on a drug regimen that requires you to take one pill a day from each of two bottles, A and B. One day you tap one pill into your palm from the A bottle and, inadvertently, two pills from the B bottle. Unfortunately the A and B pills are indistinguishable, and taking more than one B pill per day is fatal. And the pills are very expensive, so you can’t afford to throw out the handful and start over. How can you arrange to take the correct dose without wasting any pills?

	Select Click for Answer
Add a second A pill to the group and cut each pill in half: Now, regardless of their identities, if you take one half of each pill you’ll be taking the equivalent of one A pill and one B pill. Tomorrow you can take the remaining halves and then return to the normal regimen.

July 14, 2015July 26, 2015 | Puzzles

Hour of Babel

A problem from the second Balkan Mathematical Olympiad, 1985:

Of the 1985 people attending an international meeting, no one speaks more than five languages, and in any subset of three attendees, at least two speak a common language. Prove that some language is spoken by at least 200 of the attendees.

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If any two of the attendees, A and B, have no language in common, then in any trio that they belong to, the third member, C, must have a language in common with one or both of them. Now suppose that the proposition we’re asked to prove is false — that is, suppose that there are not more than 198 other attendees who speak a language that’s spoken by A or B. A and B know at most 10 languages between them, so at most there are 10 × 198 = 1980 people C who are available to complete an acceptable trio with A and B. But we’ve been told that there are 1985 – 2 = 1983 others who are able to fulfill this role. This is a contradiction, so our supposition, that no language is spoken by 200 attendees, must be false. “It is easy to forget all about the alternative case, for the odds must be overwhelming that some two of the 1985 participants will be at linguistic loggerheads,” notes Ross Honsberger in From Erdös to Kiev. “However, we still need to consider the possibility that every pair (A, B) have a language in common. Fortunately, this is a trivial case, for then each of the other 1984 people must speak one of the five or fewer languages spoken by A, for an average of at least 397 (counting A).”

Click for Answer

July 7, 2015July 19, 2015 | Puzzles

Black and White

loyd chess problem

By Sam Loyd. White to mate in two moves.

	Select Click for Answer
1. Qa1!, threatening 2. Kf7#. If 1. … Kg8 then 2. Qa8#, and if 1. … Bg8 then 2. Kxg6#. From the Detroit Free Press, 1877.

July 6, 2015July 19, 2015 | Puzzles

Cube Route

You’re planning to make the wire skeleton of a cube by arranging 12 equal lengths of wire as shown and soldering them at the corners.

It occurs to you that you might be able to simplify the job by using one or more longer lengths of wire and bending them into right angles at the cube’s corners.

If you adopt that plan, what’s the smallest number of corners where soldering will still be necessary?

	Select Click for Answer
Eight. An odd number of wires meet at each corner, so every corner will still need soldering.

June 15, 2015July 19, 2015 | Puzzles

Weighing in Verse

How can you find one bad coin among 12 using only three weighings in a pan balance? We published the classic solution in November 2013, but a clever alternative appeared in Eureka in October 1950. Here the aspiring counterfeiter is named Felix Fiddlesticks, and he’s trying to find the coin for this mother:

F set the coins out in a row
And chalked on each a letter, so,
To form the words: “F AM NOT LICKED”
(An idea in his brain had clicked).

And now his mother he’ll enjoin:

weighing in verse table

This plan will reveal the bad coin no matter which it is. For instance, if the coin marked O is heavy, then in the first two weighings the left pan will drop, and in the third weighing it will rise; no other bad coin will produce this result. The plan will also reveal reliably whether the bad coin is heavy or light.

Such cases number twenty-five,
And by F’s scheme we so contrive
No two agree in their effect,
As is with pen and patience checked:
And so the dud is found. Be as it may
It only goes to show CRIME DOES NOT PAY.

(Cedric Smith, “The Twelve Coin Problem,” The Mathematical Gazette, Vol. 89, No. 515 [July 2005], pp. 280-281. A generalized algorithm for solving such a problem for any number of coins is given by Michael Weatherfield in the same issue, pp. 275-279.)

June 14, 2015June 13, 2015 | Puzzles