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?
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.
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.)
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.
A team of volunteers have deciphered a message written by Charles Dickens in his own puzzling brand of shorthand, solving a riddle that had persisted for more than 150 years.
Apparently in 1859 the Times had mistakenly rejected an advertisement that Dickens had hoped to run during his delicate transition from the editorship of Household Words to All The Year Round. Dickens had written to the newspaper’s editor, J.T. Delane, asking him to intervene in the matter and had saved a cryptic copy of the message, possibly for legal reasons. With the passage of time the key to the author’s so-called Brachygraphy had been lost.
When scholars recently appealed for help in understanding the message, an international team of amateur solvers pooled their insights to decipher the “Tavistock Letter.” “Having the text of this letter at long last will allow scholars to learn more about Dickens’s shorthand method while gaining further insight into his life and work,” wrote Philip Palmer, curator and head of literary and historical manuscripts at Morgan Library & Museum. “We are thrilled that colleagues at the Dickens Code project have helped make this letter accessible in new ways to researchers.”
That’s not the end of it — a further puzzling page, this one from the notebooks of Dickens’ shorthand pupil Arthur Stone, still awaits solution.
(Thanks, Bill.)
From reader Derek Christie:
Each player in a game of cribbage has a hand of four cards. A single further card is turned up and serves as the fifth card in every player’s hand. Part of the game involves scoring your hand. You get points for any combination of cards that adds to 15, like 9 4 2; for two or more of any rank, like 3 3; or for any run of three or more, like Ace 2 3. The Jack, Queen, and King each score 10. Show that if a 5 has been turned up, every player must score some points.
Your eccentric uncle has dropped you into the middle of Twine Island, which is festooned with one continuous loop of twine. The twine never crosses itself, but it snakes everywhere, and the island is too hilly for you to see the whole layout at once. As a character-building exercise, your uncle offers you a million dollars if you can determine whether you’re inside the loop or outside. How can you do this?
