Black and White

lincoln chess problem

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

	Select Click for Answer
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:

	Select Click for Answer
(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?

	Select Click for Answer
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

	Select Click for Answer
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.)

Click for Answer

August 17, 2017June 1, 2018 | Puzzles

Light Work

You have 10 stacks of silver dollars, with 10 coins in each stack. The coins appear identical, but you know that all the coins in one stack are counterfeit. You know the weight of a genuine coin, and you know that a counterfeit coin weighs 1 gram less than this. How many weighings must you do to find the counterfeit stack?

	Select Click for Answer
One. Take one coin from the first stack, two from the second, and so on, and weigh this collection together. The number of grams by which the total is light is the number of the counterfeit stack.

August 16, 2017August 15, 2017 | Puzzles

Figuring

An eight-digit number contains two 1s, two 2s, two 3s, and two 4s. The 1s are separated by 1 digit, the 2s by 2 digits, the 3s by 3 digits, and the 4s by 4 digits. What is the number?

	Select Click for Answer
41312432 (or its reverse, 23421314).

August 12, 2017 | Puzzles

Black and White

shinkman chess problem

By William Anthony Shinkman, 1880. White to mate in two moves.

	Select Click for Answer
1. Rd1! and the queen will mate.

August 11, 2017August 11, 2017 | Puzzles

The Last Detail

A puzzle by R.P. Cross:

Find the last digit in the evaluation of $\sum_{n = 1}^{100}n!$

	Select Click for Answer
It would take a lot of pencils to work this out by hand, but an insight makes it easy: For any n ≥ 5, n! must have the factors 2 and 5 and hence be a multiple of 10, with zero as a final digit. So to solve the puzzle we only have to work out 1! + 2! + 3! + 4! = 33. The last digit is 3. (“A Property of Large Factorial Numbers,” Mathematical Gazette 81:490 [March 1997], 80-81.)

August 7, 2017 | Puzzles · Science & Math

Black and White

kalendovský chess problem

By Jan Kalendovský. White to mate in two moves.

	Select Click for Answer
1. Rf2! Bxf2+ 2. Kxf2#

July 24, 2017July 24, 2017 | Puzzles

Late Again

kendall/thomas railroad puzzle

A problem from P.M.H. Kendall and G.M. Thomas’ Mathematical Puzzles for the Connoisseur, 1962: A road runs parallel to a railway until it bends to cross it, as shown. A man normally cycles to work along the road at a constant speed of 12 mph, and when he reaches the crossing he’s normally overtaken by a train traveling in the same direction. One day he was 25 minutes late for work and found that the train passed him 6 miles before the crossing. What was the speed of the train?

	Select Click for Answer
Suppose the man normally reaches the crossing, B, at 8 a.m. Then on a typical day he’d reach A, 6 miles before the crossing, at 7:30 a.m. When the man is 25 minutes late, he’ll reach B at 8:25 a.m. and A at 7:55 a.m. Hence the train takes 5 minutes to travel the 6 miles from A to B. Therefore its speed is 72 mph.

July 15, 2017July 14, 2017 | Puzzles