Readouts

sallows readouts puzzle

A puzzle by Lee Sallows. In this readout from a computer-driven electronic display, the digits in the fifth row have been obscured. What are they?

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They’re the same as the numbers in the first row. Added colors make it easier to see that each number in row n counts how many segments of each type (color) occur in row n – 1. Remarkably, the loop closes after four cycles. (Thanks, Lee.)

May 16, 2022May 15, 2022 | Puzzles

Staffing

A problem from Crux Mathematicorum, April 2006:

A group of people must be formed into committees. Show that the number of possible committees that can be formed with an odd number of members is the same as the number that can be formed with an even number of members. (Assume that a committee with no members and one that includes everyone are both allowed.)

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This solution is by Geneviève Lalonde. Call a committee an even committee if it has an even number of members and an odd committee if it has an odd number of members. If the total number of people in the group is odd, then for each even committee the total number of remaining people is odd (and thus constitutes an odd committee). And vice versa: For each odd committee, the number of people who aren’t in that committee is even and constitutes an even committee. So there’s a one-to-one correspondence between the even and odd committees, and their numbers must be equal. If the total number of people in the group is even, then consider one person (call him John). For each even committee C that doesn’t contain John, there’s a corresponding odd committee made up of everyone who’s not in C except for John. And for each even committee C that contains John, there’s a corresponding odd committee made up of John plus everyone who’s not in C. So again there’s a one-to-one correspondence between the even and odd committees, and we conclude that their numbers are equal.

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May 5, 2022May 4, 2022 | Puzzles

Black and White

taylor chess problem

By J. Paul Taylor. White to mate in two moves.

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1. Be4! and mate must follow. From Elementary Chess Problems, 1880.

April 29, 2022April 28, 2022 | Puzzles

Can Do

A puzzle by Sam Loyd:

John the milkman has two 10-gallon cans full of milk. Two customers have a 5- and a 4-quart measure and want 2 quarts put into each measure. How can he accomplish this?

“It is a juggling trick pure and simple, devoid of trick or device, but it calls for much cleverness to get two exact quarts of milk into those measures employing no receptacles of any kind except the two measures and the two full cans.”

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Call one of the 10-gallon cans A and the other B. Fill the 5-quart pail from can A. Pour the 5-quart pail into the 4-quart pail. Empty the 4-quart pail into can A. Pour the 5-quart pail into the 4-quart pail. Fill the 5-quart pail from can A. Fill the 4-quart pail from the 5-quart pail. Empty the 4-quart pail into can A. Fill the 4-quart pail from can B. Pour the 4-quart pail into can A. This fills can A and leaves 2 quarts in the 4-quart pail. “Thus the milkman has supplied each of his customers with exactly two quarts of milk and solved his perplexing problem.” From Loyd’s Cyclopedia of Puzzles, 1914. See also Well Done.

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April 28, 2022April 27, 2022 | Puzzles

Perspectives

eureka elevation puzzle

A puzzle from Eureka, March 1957:

These drawings depict an object as seen from above and from the front. What does it look like when viewed from the side? The object has no curved surfaces, and all hidden edges in such drawings are represented by dotted lines.

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(From the Eureka Digital Archive, published under a Creative Commons license.)

April 22, 2022April 21, 2022 | Puzzles

An Observant Anthropologist

A puzzle from the 1998 Moscow Mathematical Olympiad, via Peter Winkler’s excellent Mathematical Puzzles, 2021:

An anthropologist is surrounded by a circle of natives. Each native either always lies or always tells the truth. The anthropologist asks each native whether the native to his right is a liar or a truth teller. From their answers, she’s able to deduce the fraction of the circle who are liars. What is the fraction?

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The key is to notice that if all the liars were changed to truth-tellers and vice versa, none of the answers would change. So if the fraction x can be determined, it must have the property that x = 1 – x, and x must be 1/2.

April 20, 2022April 19, 2022 | Puzzles

Two Weighings

A problem from the Leningrad Mathematical Olympiad: You have a set of 101 coins, and you know that it contains one counterfeit coin X. The 100 genuine coins all have the same weight, which is different from that of X. Using only two weighings in an equal-arm balance, how can you determine whether X is heavier or lighter than the genuine coins?

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Divide the coins arbitrarily into piles of 30, 30, and 41 coins, calling these respectively A, B, and C. Now weigh A against B. If these balance, then X must be in C; combine A and B and weigh 41 of these coins against the 41 coins in C and you’ll find out whether X is heavy or light. If A and B don’t balance, then one of those groups contains the bad coin. For definiteness, suppose that A goes down in the weighing. That means either that A contains the bad coin and it’s heavy or that B contains the bad coin and it’s light. So determining where X is will also tell us its relative weight. Divide A in half and weigh these halves against each other. If they balance then the coins are all good and X must be in B. If they don’t balance then X is in A. (Obviously the same plan will work with certain pile sizes other than 30, 30, and 41, but that arrangement will do the job.)

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April 9, 2022April 8, 2022 | Puzzles

A Late Mystery

In Lloyd C. Douglas’ 1929 novel Magnificent Obsession, a doctor dies of a heart attack, leaving behind a journal written in cipher. The first page is shown here. Can you read it?

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It’s a rail fence cipher. To see it explained, search the text of the book for the sentence “Then a fresh idea occurred to him.” (To understand its import, you can read the full text or just the novel’s Wikipedia page, especially the section “Dr. Hudson’s Secret Journal.”)

April 5, 2022April 4, 2022 | Literature · Puzzles

Where Did Nigel Go?

A puzzle from the excellent Riddler feature at FiveThirtyEight, via Oliver Roeder’s 2018 collection The Riddler:

Your eccentric friend Nigel flies from Heathrow to an airport somewhere in the 48 contiguous states, then hires a car and drives around the country, touching the Atlantic and Pacific Oceans and the Gulf of Mexico, then returns to the airport at which he started and flies home. If he crossed the Ohio River once, the Missouri River twice, the Mississippi River three times, and the Continental Divide four times, then there’s one state that we can say for certain that he visited on his trip. What is it?

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Image: Wikimedia Commons Minnesota. In order to cross a river an odd number of times and still return to the original airport, Nigel must “go around” the river somehow. That’s not hard with the Ohio, but the Mississippi is enormous and practically divides the country. So in order to cross it three times and return to his airport, at some point he’d have had to go around the source at Lake Itasca, Minnesota.

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April 1, 2022March 31, 2022 | Puzzles

Trainspotting

A puzzle from James F. Fixx’s More Games for the Superintelligent, 1976:

A man who likes trains walks occasionally to a nearby railroad track and waits for one to go by. Afterward he notes whether he saw a passenger train or a freight. After several years his notes show that 90 percent of the trains he’s seen have been passenger trains. One day he meets an official of the railroad and is surprised to learn that the passenger and freight trains on this line are precisely equal in number. If the man timed his trips to the track at random, why did he see such a disproportionate number of passenger trains?

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All the trains run hourly, but the freight trains follow six minutes after the passenger trains. So if the man arrives at the track at random, his odds of arriving after a freight train and before a passenger train are 9 to 1; for 54 out of every 60 minutes, a passenger train will be the next to pass.

March 31, 2022March 30, 2022 | Puzzles