(Bloomberg Opinion) -- In this golden age of printing money, it seems like a great time to enjoy a coin-weighing puzzle — a type of Conundrum that has been gaining currency since the birth of cash itself. This one features enough change to put Scrooge McDuck in the mood for a swim:
You have 20 stacks of 20 identical-looking golden coins — but all the coins in one of the stacks are counterfeit! (The coins in the other 19 stacks are real.) You know that counterfeit coins are heavier than genuine ones. And you have an electronic scale that can determine the exact weight of any number of coins, but its batteries are almost fully drained.
Can you figure out which stack has the counterfeit coins in just three weighings?
To start solving this puzzle, which I debuted last week as part of a series I put together for this year’s Regeneron International Science and Engineering Fair,(1)you might use a strategy from a previous column: simplification. When a puzzle is so big as to seem unfathomable, try to shrink it down by any means necessary:
1. Could you find the counterfeits if you knew how much real and fake coins weighed?
2. What if there were only a couple of stacks?
If you figure it out this weighty challenge — or even make partial progress — please let me know at firstname.lastname@example.org before midnight Eastern time on Wednesday, May 27. (If you get stuck, there’ll be a hint announced in Bloomberg Opinion Today on Tuesday, May 26. Sign up here.)
Last Week’s Conundrum
The code was the word, with a pair of cryptograms. First, we solved a simple Caesar shift as a warm-up. The encoding just shifted each letter forward by two – “A” to “C”; “B” to “D,” and so forth – and the punny message was:
I’M A CAESAR CIPHER!
OH YEAH? THEN WHERE ARE YOUR LETTUCE AND CROUTONS??
Then we moved on to the main event, which was a mysterious cipher made up of numbers. Some of the numbers were integers, while others had a decimal point, followed by a tenths’ place digit.
The first line, for example, read:
11.4 / .1 3 7.1 2.2 11 16 / 17.4 18.2 9 15 ?
While the numbers may have looked random at first, astute solvers noticed some structure: the tenths’-place digit, when it appeared, was always between 1 and 5, and the largest number overall was 21.
5 + 21 = 26, which happens to be the number of letters in the English alphabet. There are 5 vowels and 21 consonants, so we might guess that the post-decimal digits stood for vowels (“.1” for “A”, “.2” for “E”, and so forth) while the pre-decimal digits represented consonants (“1” for “B”, “2” for “C”…).(2)
But no number has multiple decimal points — so for that encoding to work out, it would have to be the case that none of the words in the message have vowels next to each other.
Is that possible? Indeed it is. Here’s the message, fully decoded:
NO ADJACENT VOWELS?
IT HAPPENS THAT MANY WORDS HAVE THIS NOTEWORTHY PROPERTY!
NICE JOB SOLVING
SEND IN A FAVORITE SUCH WORD AS THE "ANSWER" TO THIS PUZZLE
Both of these puzzles highlight a trick cryptogram constructors often play: using unusual words (like “croutons”) or word arrangements (no adjacent vowels). This provides a sort of camouflage that makes it harder to guess the message just from its letter patterns.
Notwithstanding,(3) 30 solvers cracked the code. Jason Hardy was first, and suggested the highly appropriate word "CIPHERTEXT"; a few minutes later, Elizabeth Sibert sent in the fully vowel-free “SYZYGY.”(4) Anna Collins came up with the longest word we received, "HONORIFICABILITUDINITATIBUS," just barely beating out Edgar Ruiz Guzman (“OTORHINOLARYNGOLOGIST”) and Winston Luo (“MAGNETOHYDRODYNAMICS”). Zoz pointed out that “AGESILOCHUS” — the chief magistrate of the Rhodians — has no adjacent vowels yet still has the rare property that each vowel appears exactly once and in order! And Mara Evans, to my great pleasure, pointed out that “CONUNDRUM” itself fits the pattern — an elegant tie-in that I myself had missed.
The Bonus Round
The Sudoku sensation that’s sweeping the nation (hat tip: Ben Chartock); escape rooms from home; club trivia from anywhere. Birdsong Hero (hat tip: ISEF); mathematical catharsis in solitary; the best way to scramble one of those 15 puzzles; and “The Artist of the Unbreakable Code.” Use marbles to play a musical instrument (hat tip: Adam Rosenfield), or just set up a really, really long marble run. And inquiring minds want to know: how did the graduate student cut that knot?
(1) Typically, the Fair is held in-person, but this year it was online. I've been involved in the Fair since competing there as a student in 2005. Most years, I serve as co-chair of Mathematics category judging; as of a couple years ago, I also serve on the National Leadership Council of Society for Science & the Public, the nonprofit organization that runs the event.
(2) I got the idea for this code from a similar cipher featured in a 1990s book series I read as a kid – Cyber.kdz, which follows a group of teenagers who fight crime over the Internet. I had a somewhat unusual childhood.
(3) Another word without adjacent vowels!
(4) Samuel Castillo got close to vowel-free with “CHRYSLER.” (And several people submitted the fully decoded ciphertext but didn't realize the message itself was an instruction.) Mad props to everyone who submitted their answers using the same cipher encoding.
This column does not necessarily reflect the opinion of the editorial board or Bloomberg LP and its owners.
Scott Duke Kominers is the MBA Class of 1960 Associate Professor of Business Administration at Harvard Business School, and a faculty affiliate of the Harvard Department of Economics. Previously, he was a junior fellow at the Harvard Society of Fellows and the inaugural research scholar at the Becker Friedman Institute for Research in Economics at the University of Chicago.
©2020 Bloomberg L.P.