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?

	Select Click for Answer
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.)

Click for Answer

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?

	Select Click for Answer
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?

	Select Click for Answer
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.

Click for Answer

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?

	Select Click for Answer
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

Triples

A brainteaser from the Soviet science magazine Kvant, via Quantum, January/February 1991:

Bobby found the sum of three consecutive integers, then of the next three consecutive integers, then multiplied these two sums together. Could the product have been 111,111,111?

	Select Click for Answer
No. One of the two groups must contain one even number and two odd ones. The sum of these three numbers must be even, so the product of the two sums must be even as well.

March 30, 2022March 29, 2022 | Puzzles

Black and White

carpenter chess problem

By George E. Carpenter. White to mate in two moves.

	Select Click for Answer
1. Bd7! and the queen will mate. From Chess Problems, 1886.

March 29, 2022March 28, 2022 | Puzzles

Line Work

bilinski problem

Robert Bilinski proposed this problem in the April 2006 issue of Crux Mathematicorum. On square ABCD, two equilateral triangles are constructed, ABE internally and BCF externally, as shown. Prove that D, E, and F are collinear.

	Select Click for Answer
This solution is by Titu Zvonaru. Triangle ADE is isosceles (because AD and AE are the same length). And ∠DAE = 90° – 60° = 30°. Similarly, triangle EBF is isosceles (because BE and BF are the same length). And ∠EBF = 60° + 30° = 90°. So $\displaystyle \angle \mathit{DEF} = \angle \mathit{DEA} + \angle \mathit{AEB} + \angle \mathit{BEF} \\\\ = \frac{180^{\circ} - 30^{\circ}}{2} + 60^{\circ} + \frac{180^{\circ} - 90^{\circ}}{2} = 180^{\circ}.$ Hence D, E, and F are collinear. 04/03/2022 UPDATE: Reader Peter Smith sent this alternative proof, which doesn’t require the calculation of angles: AECFB is a symmetrical pentagon (containing two equilateral triangles); therefore EC is parallel to AF. It follows that ∠DCE is equal to ∠BAF. By symmetry about a vertical axis, ∠CDE is equal to ∠DCE. By symmetry about a horizontal axis, ∠CDF is equal to ∠BAF. Therefore ∠CDE is equal to ∠CDF and DEF must be collinear. (Thanks, Peter.)

Click for Answer

March 26, 2022April 3, 2022 | Puzzles

A Century-Old Ghost

What does this mean?

PMVEB DWXZA XKKHQ RNFMJ VATAD YRJON FGRKD TSVWF TCRWC
RLKRW ZCNBC FCONW FNOEZ QLEJB HUVLY OPFIN ZMHWC RZULG
BGXLA GLZCZ GWXAH RITNW ZCQYR KFWVL CYGZE NQRNI JFEPS
RWCZV TIZAQ LVEYI QVZMO RWQHL CBWZL HBPEF PROVE ZFWGZ
RWLJG RANKZ ECVAW TRLBW URVSP KXWFR DOHAR RSRJJ NFJRT
AXIJU RCRCP EVPGR ORAXA EFIQV QNIRV CNMTE LKHDC RXISG
RGNLE RAFXO VBOBU CUXGT UEVBR ZSZSO RZIHE FVWCN OBPED
ZGRAN IFIZD MFZEZ OVCJS DPRJH HVCRG IPCIF WHUKB NHKTV
IVONS TNADX UNQDY PERRB PNSOR ZCLRE MLZKR YZNMN PJMQB
RMJZL IKEFV CDRRN RHENC TKAXZ ESKDR GZCXD SQFGD CXSTE
ZCZNI GFHGN ESUNR LYKDA AVAVX QYVEQ FMWET ZODJY RMLZJ
QOBQ-

No one knows. Cryptologist Louis Kruh discovered it in the New York Public Library’s rare book room in 1993 among some old material from the U.S. Army Signal School. In 1915 first lieutenant Joseph O. Mauborgne had created what he believed was a more secure cipher than the ones currently in use, and had offered this challenge to see if his colleagues could break it. Kruh found no solution in the archive, and he published it in both The Cryptogram and Cryptologia, inviting their readers to try their hands at it. As far as I know, none succeeded.

Mauborgne described it as a “a simple, single-letter substitution cipher adapted to military use.” He invited the director of the Army Signal School to place it on a bulletin board and allow the officers there to work on it for three months and then to post the solution “to show why the standard method of attacking a substitution cipher fails in this case.” “If any attack upon this cipher is successful, I shall be glad to hear of it,” he wrote.

Kruh, who died in 2010, noted that “it was probably solved or otherwise deemed unsuitable for use because there is no knowledge of a new cipher being adopted by the Army around that time.” If a solution was found, I don’t believe anyone alive today knows what it is.

(Louis Kruh, “A 77-Year-Old Challenge Cipher,” Cryptologia 17:2 [April 1993], 172-174.)

March 14, 2022March 13, 2022 | Puzzles · Science & Math

Setting the Date

A romantic puzzle from Albert H. Beiler’s Recreations in the Theory of Numbers:

An ardent swain said to his lady love, some years ago, ‘Once when a week ago last Tuesday was tomorrow, you said, “When a day just two fortnights hence will be yesterday, let us get married as it will be just this day next month.” Now sweetheart, we have waited just a fortnight so as it is now the second of the month let us figure out our wedding day.’

Beiler’s book came out in 1964, so he gives the answer Tuesday, March 17, 1936 — the couple are speaking on March 2, discussing a conversation they had on February 17. Obviously the answer is not unique — “Tuesday, March 17, 1908, is another solution but then the swain would not be very young.” Basically we need a leap year in which March 17 falls on a Tuesday. Beiler finds these occur also in 1964 and 1992 — and one did in 2020 as well.

February 24, 2022February 23, 2022 | Puzzles

Black and White

brown chess problem

By John Brown. White to mate in two moves.

	Select Click for Answer
1. Qe7! and the queen or bishop will mate. From Chess Strategy, 1865.

February 20, 2022 | Puzzles