A Mating Machine

elkies chess puzzle

A “no-brainer” by Harvard mathematician (and chess master) Noam Elkies. White has choices as to how to play, but no matter how he proceeds, he’ll wind up mating Black on the seventh move.

(From Tim Krabbé’s chess diary.)

November 10, 2017 | Puzzles

Black and White

tarrant chess problem

By Thomas Tarrant. White to mate in two moves.

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1. Nf5! and 2. Rb5 will mate. From the Westminster Papers, 1877.

November 2, 2017November 1, 2017 | Puzzles

Coverup

A problem from the British Columbia Colleges Senior High School Contest for 2000:

If I place a 6 cm × 6 cm square on a triangle, I can cover up to 60% of the triangle. If I place the triangle on the square, I can cover up to 2/3 of the square. What is the area, in cm², of the triangle?

(a) 22 4/5
(b) 24
(c) 36
(d) 40
(e) 60

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The answer is (d). The key is to see that these measures are one and the same: The triangle overlaps the square by the same amount that the square “underlaps” the triangle, so these areas are maximized at the same moment. If A is the area of the triangle, then 0.6A = 2/3 × 36, and A = 40 cm². From “The Skoliad Corner,” Crux Mathematicorum, December 2000.

October 18, 2017October 17, 2017 | Puzzles

Black and White

shinkman chess problem

“We now give what is acknowledged to be the finest two-move problem extant,” wrote J.H. Blackburne in the Strand in 1908. “It is by the American expert, W.A. Shinkman, and is also claimed by G.E. Carpenter, a fellow countryman of his. Here we have not only a difficult key-move, but also beauty of theme and artistic construction, the three essential qualities necessary to a perfect problem.”

White to mate in two moves.

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“It is solved by B to R 4th [that is, 1. Ba4], moving the B from a strong position, where it defended a Kt, to, apparently, a useless square. In reply, Black has a choice of five moves”: 1. … f5 2. Qg8# 1. … f6 2. Ndc7# 1. … e4 2. Qxe4# 1. … Kxd5 2. Bb3# 1. … d6 2. Nbc7# “Truly a marvellous composition.” (“Bridge and Chess Problems,” Strand, March 1908.)

October 12, 2017October 11, 2017 | Puzzles

The Fifth Element

pmej matrix puzzle

Charles W. Trigg offered this puzzle in the Fall 1977 issue of Pi Mu Epsilon Journal (PDF):

This square array contains the first 25 positive integers. Choose five, no two from the same row or column, so that the largest of the five elements is as small as possible, and justify your choice.

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Sidney Penner of Bronx Community College of CUNY, found the solution: 1, 3, 6, 8, 11. In the array, crossing out all the numbers higher than 11 eliminates all the numbers in the third row except 8. Choosing 8 means we have to forswear every other number in the fifth column, which leaves only one possibility in the fourth row, 3. And choosing 3 eliminates every other candidate in the first column, so the only available entry in the first row is 11. At the request of problem editor Leon Bankoff, Jeanette Bickley ran the following BASIC program on a DEC PDP 11/70 computer, and it confirms Penner’s solution:

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October 11, 2017October 10, 2017 | Puzzles

The Impossible Puzzle

Dutch mathematician Hans Freudenthal proposed this puzzle in 1969 — at first it appears impossible because so little information is given.

X and Y are two different whole numbers greater than 1. Y is greater than X, and their sum is no greater than 100. S and P are two logicians; S knows the sum X + Y, and P knows the product X × Y. S and P both reason perfectly, and both know everything I’ve just told you.

S says, “P does not know X and Y.”
P says, “Now I know X and Y.”
S says, “Now I also know X and Y.”

What are X and Y?

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X = 4 and Y = 13. S knows the sum is 17, and P knows the product is 52. S makes a list of all the addends that could produce 17 and fulfill the other conditions above: 2 + 15 (product: 30) 3 + 14 (product: 42) 4 + 13 (product: 52) 5 + 12 (product: 60) 6 + 11 (product: 66) 7 + 10 (product: 70) 8 + 9 (product: 72) S knows that P is thinking of one of the products listed above; she can say that P doesn’t know X and Y because each of these products has multiple solutions (e.g. 30 = 2 × 15 = 3 × 10). P has been building his own list — in his case, he’s listing pairs of factors that produce 52. It’s a short list — there are only two possibilities that meet the conditions: 2 × 26 4 × 13 When S makes her first announcement, P understands that her candidate list is like the one above — it’s one in which all of the products are ambiguous. And this eliminates the first candidate on P’s list. If X = 2 and Y = 26, then their product would be 52, as P expects, but S would know that their sum was 28, and she couldn’t then claim that P didn’t know the numbers — her list of suspects would look like this: 2 + 26 (product: 52) 3 + 25 (product: 75) 4 + 24 (product: 96) 5 + 23 (product: 115) … That fourth item, X = 5 and Y = 23, gives a product of 115, which has no other set of factors that meet the requirements. So if 28 were the sum, S couldn’t have declared so confidently that P was still in the dark. Realizing this, P can eliminate 2 × 26 from his list, leaving only one possibility: X is 4 and Y is 13. He says, “Now I know X and Y.” And by reconstructing his thinking, S can determine these values herself. (Thanks, Jon.)

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September 24, 2017 | Puzzles · Science & Math

Black and White

lincoln chess problem

By Robert A. Lincoln. White to mate in two moves.

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1. Ka5! Kxa2 2. Kb4#

September 21, 2017 | Puzzles

Four-Sided Story

University of Ljubljana electrical engineer Izidor Hafner devised this maze on a tetrahedron, presented as an unfolded plane plan. Can you find a path from one dot to the other? To do so you’ll have to fold up the figure in your head:

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(From Crux Mathematicorum, December 2000. See Hafner’s Wolfram Demonstrations page for mazes on cylinders, polyhedra, tori, Moebius strips, Riemann surfaces, and more.)

September 19, 2017September 18, 2017 | Puzzles

Black and White

piccinini chess puzzle

By Augusto Piccinini. Imagine that the board is a vertical cylinder, that is, that the a-file and the h-file are joined so that pieces can move across the border. How can White mate Black in two moves?

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1. a5! leaves the black king with no moves. Black must move the pawn, whereupon White captures it, giving mate. (If Black moves the pawn two squares, White can capture it en passant, giving the same result.) From Rivista Scacchistica Italiana, 1907.

September 1, 2017 | Puzzles

Incognito

What three digits are represented by X, Y, and Z in this addition problem?

incognito

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The Xs in the left column produce a 1-digit sum, so X = 1, 2, 3, or 4. But looking at the right column, X is the sum of Y and Z, which are distinct numbers, so the smallest possible value of X is 1 + 2 = 3. So X must be either 3 or 4. That means (looking at the left column again) that Y must be at least 6. But Y and Z are summed in both the center and the right columns, producing different digits. So Y and Z must total at least 11, and that means that a 1 is being carried into the total X + X in the left column, producing a total for that column (and a value for Y) of either 3 + 3 + 1 = 7 or 4 + 4 + 1 = 9. The fact that we’re carrying a 1 into the center column shows us that X + 1 = Z. We’ve established that X is 3 or 4, so Z must be 4 or 5. So: X = 3 or 4 Y = 7 or 9 Z = 4 or 5 A bit of trial and error shows that X = 4, Y = 9, Z = 5. 08/18/2017 UPDATE: There’s a simple solution that I overlooked: The tens place shows Z + Y = Z. So either Y is 0 or Y is 9 and we’ve carried a 1 from the ones place. And Y can’t be 0 because it’s the first digit of the solution. So Y is 9. The hundreds place shows X + X = 9. That’s possible only if X is 4 and we’ve carried a 1 from the tens column. And the ones place shows Y + Z = X; that is, 9 + Z = 4. So Z is 5. Also, my argument that X can’t be 1 or 2 isn’t valid — the X in the first column could be the ones digit of some two-digit sum. Thanks to everyone who wrote in about this; sorry I couldn’t reply to each of you individually. 06/01/2018 Yet another solution: From the first column we know that X < Y, since there is no 1 carried forward. This means the third column is effectively Y + Z = X + 10 (a 1 has to be carried forward, per the observation above). Now the second column gives this equation: Z + Y + 1 = Z + 10 (again a 1 must be carried forward, since Y can’t be zero). Hence the first column gives the equation: X + X + 1 = Y. Solving the three equations gives the three variables. (Thanks, Tom.)

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August 17, 2017June 1, 2018 | Puzzles