ISEF Conundrums 2021 - NET
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ISEF Conundrums – 2021
Scott Duke Kominers (ISEF’05)
MONDAY
We started with a bit of a warm-up, asking: Can you crack this code?
18.2 9 2.4 10.2 / 16.4 / .3 15.2 4 ! / 15.2 11 3 / .5 15 / .1
/ 4.1 17.4 14.3 16.2 / 15 2 6.4 9.1 14 / .4 14 / .3 11 11.4 17.1
16.4 14 / .1 15 / 16 6.2 / ‘‘ .1 11 15 18.2 14 " / 16.4 / 16 6.3
15 / 12.5 21 21 9.2 ; / 1.4 11.5 15 / .3 4 / 11.4 / .1 3 7.1 2.2
11 16 / 17.4 18.2 9 15 / .3 11 / 11.1 10.2 !
This was a rather unusual-looking cipher: it had lots of different numbers in it, some
with decimals and some without. Frequency analyses of the different numbers didn’t match
up with anything that looked like English (or any other language, for that matter)—but that
was because some of the numbers were actually standing in for two letters, rather than one!
Each number after the decimal point represented one of the five English-language vowels—
A, E, I, O, or U (indicated in alphabetical order); the other numbers stood for the consonants
(again, in order). So for example, “18.2” at the beginning stood for the letters “WE,” that is,
the 18th consonant followed by the 2nd vowel. (Notably, for this encoding scheme to work,
the entire message had to avoid having adjacent vowels, otherwise we would have ended up
with two decimal points in a single number.)
Fully decoded, the phrase read as follows:
WELCOME TO ISEF! SEND US A FAVORITE SCHOLAR OR INNO-
VATOR AS THE “ANSWER” TO THIS PUZZLE; BONUS IF NO AD-
JACENT VOWELS IN NAME!
Becca Barbera, Maria Bayder, Peter Jay Corros, Omer Eyal, Riya Gupta, David Hol-
lander, Lazar Ilic, Nikhil Iyer, Rishab Jain, Layali Khatib, Adelina Kildeeva, Sydney Kr-
usch, Zoe Lakkis, Matthew Lee, Franklin López, Natalie McGee, Mayu Nakano, Catherine
Oikonomou, Kosuke Oya, Joonwoo Park, AnaMaria Perez, Federico Runco, Yasutaka Toku-
maru, Franklin Wang, Ellen Xu, and Alexander Zhang managed to decode the message, and
sent in the following favorite scholars and innovators (not listed in order): Ada Lovelace, Alan
Turing, Andrew Ng, Benjamin Franklin, Elizabeth Feldman, Emil Artin, Erwin Schrödinger,
Frances Arnold, James Naismith, Leonardo da Vinci, Linus Benedict Torvalds, Maryam
For more Conundrums, please sign up for the weekly email list here).
1Mirzakhani, Nikola Tesla, Richard Feynman, Rosamund Franklin, Sigmund Freud, Stephen Hawking, Steve Jobs, and Tycho Brahe. (Special kudos to Ilic, Iyer, Oikonomou, Tokumaru, and Zhang, who encoded their an- swer submissions using the same cipher!)
TUESDAY
This year there are 1834 students at ISEF, across 1471 projects—from 64 countries,
regions, and territories.
That’s a lot of students! But can you prove that at the end of ISEF week, there will be
at least two of them who know the same number of other students at the Fair?
Now suppose that exactly two students somehow manage to meet all the others (including
each other). Then they each know 1833 other students at the Fair. Would there still be two
students other than those two who know the same number of other students?
The basic version of this puzzle is well-known, and I’m not sure where it originally
appeared. But I first learned it from Lagoon Books’s Mind-Bending Classic Logic Puzzles.
The trick, as Becca Barbera, Maria Bayder, Maddy Christensen, Peter Jay Corros, Riya
Gupta, David Hollander, Lazar Ilic, Nikhil Iyer, Rishab Jain, Adelina Kildeeva, Yuhao Lu,
Natalie McGee, Catherine Oikonomou, Joonwoo Park, Marco Ruocco, Franklin Wang, Ellen
Xu, Sama Zahran, Jason Zhang, and Alexander Zhang figured out, is to compare the total
number of students with the total number of numbers of students that different students can
know (now there’s a tongue-twister for you!).
There are 1834 students, and at first it looks like there might be 1834 different possible
counts. But as Kildeeva explained in her solution:
“The maximum possible is to get to know everyone, that is, to have 1833
acquaintances. Suppose this is possible and everyone knows a different
number of students. Then the only possible option is for different finalists
to know
0, 1, 2, . . . , 1832, 1833
students. But if someone knows all the participants (1833 students), then
there is no student who does not know anyone (0 students).”
Thus, by mathematical a principle called the pigeonhole principle (if you have N pigeons to
fit into N − 1 pigeonholes, then at least two pigeons must be in the same hole), there must
be at least two students who know the same number of other students.
For the second part, a version of the same argument worked: with two students knowing
all of the other students, there cannot be any students who know 0 or 1 other. So then the
remaining options for the remaining 1832 students are
2, . . . , 1832, 1833;
if one of them knows 1833, then none of them can know just 2, so once again there must be
two students who know the same number of other students.1
1We had intended to simplify the problem slightly by imposing the constraint that exactly two students
somehow manage to meet all the others, which would also eliminate the option of 1833; as nearly all the
solvers noticed, this was unnecessary.WEDNESDAY
This Conundrum was something of a creative challenge:
You might have played the game “24,” in which you’re given a set of four integers and have
to construct the number 24 out of them using mathematical operations. (So for example,
“1, 3, 3, 6” would have the solution 24 = 3 · (1 + 6) + 3.)
The goal was to do the same sort of thing with the number 2021, starting from factors
of 100 (since Society for Science is celebrating its centennial!).
To make the problem interesting, solvers could use as many or as few factors of 100 as
they wanted, and could use them with repetition. Plus any and all mathematical operations
were fair game!2
We challenged solvers first to construct 2021 using the smallest number of factors of 100.
(For constructions using the same number of factors, we broke ties in favor of the ones that
for which the largest factor used is the smallest.)
And we also asked to see the most creative or mathematically elegant constructions
solvers could come up with.
We received submissions from Jason Apostol, Becca Barbera, Maria Bayder, Maddy
Christensen, Peter Jay Corros, Omer Eyal, Chris Gould, Lazar Ilic, Adelina Kildeeva, Syd-
ney Krusch, Sydney Krusch, Arati Kulkarni, Zoe Lakkis, Yuhao Lu, Mayu Nakano, Jennie
Orr Thomas, Joonwoo Park, Richard Rhine, Franklin Wang, Ellen Xu, Enqi Yang, Jason
Zhang, and Alexander Zhang.
Franklin Wang managed to construct 2021 using the smallest number of factors —just
two copies of the factor 1, and a lot of iterated operators:
lp m
σ (σ (σ (σ (σ (dexp (σ (σ (dsinh (1)e)))e))))) + exp (σ (dsinh (1)e)) .
Here, σ(·) is the sum of divisors function, and d·e is the “least integer greater than” (or
“ceiling”) function.
We really enjoyed reading all the other constructions, as well! Here are a few of our
favorites:
• Ellen Xu’s, with many 10s:
2021 = 10 · 10 · log(10 · 10) · 10 + log(10 · 10) · 10 + 10/10.
• Chris Gould’s and Jennie Orr Thomas’s, both of which made use of many 2s and
5s:
2021 = 2 · (22·5 − 1) − 52 ; 2021 = 5(5 · 22 )2 + 5 · 22 + 2/2.
• Adelina Kildeeva’s, using each factor exactly once:
2021 = 1 + (2 + 4) · 5 − 10 + (50 − 25) · (100 − 20).
2That said, we reserved the right to disqualify operations that are particularly artificial, such as ones
that identify specific elements of sequences for which 2021 is simply the first element.• Zoe Lakkis’s spectacular summation:
∞ n
X 1
2021 = 1− .
n=0
20(102 + 1) + 1
• Joonwoo Park’s, with many square roots:
√ √ √ √ √ √ √
2021 = 10 · 20 · 50 + 1 · 100 · 4 + 1.
• And finally, Peter Jay Corros’s, which is especially visually pleasing:
2021 = (1 + 1)1+1+1+1+1+1+1+1+1+1+1 − (1 + 1 + 1)1+1+1 .THURSDAY
An old classic for #ThrowbackThursday:
Professor Perplex has been called in at the last minute to tally the results of a student
government election. But unfortunately he hasn’t been following the campaigns at all, and so
he doesn’t know how many candidates there are, or how many total votes were cast. Moreover,
most of his thinking is dedicated to complex mathematical theorems, so he doesn’t have much
focus to spare on the election: he can only remember one name and one number at any given
moment. (Plus the number can’t be that big—several or even 100 times the size of the size
of the student body is fine, but Professor Perplex can’t remember a number large enough to
encode the details of all the ballots he sees.)
Each ballot just has a single name on it, representing a vote for that candidate. Professor
Perplex will be passed these ballots one at a time. While he has a ballot, he can adjust the
name and number he has in memory if he wants. But once he’s done with the ballot, it is
put in a lockbox to prevent accidental double-counting, and he can’t access it again.
Can you find a strategy that guarantees Professor Perplex can always identify the winner
of the election, so long as the top candidate has more than 50% of the vote?
This is my favorite mathematical brainteaser of all-time, which I learned from Peter
Winkler’s Mathematical Puzzles: a Connoisseur’s Collection.
Becca Barbera, Maria Bayder, Lazar Ilic, Joonwoo Park, Ellen Xu, and Alexander Zhang
figured out the correct solution.
Zhang described the approach as follows:
You start by remembering the first candidate and the number 1. Go through
each ballot one by one. If the candidate matches the one in your memory,
add 1 to the number you remember; otherwise, subtract 1. If your number
reaches 0, it means the candidate in your memory no longer has more than
50% of the votes so far, because there were an equal number of votes for
and not for them. [So you restart the process with the next candidate’s
name.]
Since we know the winning candidate has over 50% of the vote, we know that they must
be the candidate in memory at the end.
(A second way to think about this is that so long as the winner has over 50% of the
vote, you will always “count” the winner more than all of the other candidates combined,
which means that at the very end you must have the winner’s name in your memory, with
a positive count.)FRIDAY
To wrap up the week, we presented an acrostic puzzle in which we imagined famous
scientists taking to social media to announce some of their ideas. (These are all imaginary,
of course—any overlap with actual social media accounts is purely accidental. And we took
a bit of liberty with the way in which the scientists and their discoveries are described, in
order to make for interesting clues.)
The task was to identify the scientists and figure out how to fit their names into the grid
below. (The clues were not in the same order as the names appear in the grid, but we filled
in one to get you started.)
After doing that, you could read out a clue to the identity of one additional scientist
whom I find particularly inspiring. Solvers were instructed to send in that person’s name as
their answer.
• @AstroMath: OMG I just figured out an incredibly efficient way to stack oranges!
I’m pretty sure it’s optimal!!!
• @BetaExpert: #ParityIsNotConserved
• @BigHairPhys: My new theory is so #massive it might distort space-time!
• @computer: The trick to getting to and from space safely is parabola math! =
Katherine JOHNSON
• @dyNamite: Planning on endowing a prize for “benefit to humankind.” RT with
category suggestions, please
• @grace: So glad I was early enough to the computer revolution to be able to get my
first name as my username!
• @HavanaEpi: Watch out for mosquitoes! q They carry #yellowfever
p √
• @HypergeometricMath: I just computed 1 + 2 1 + 3 1 + · · ·.
• @MetaModuli: Want to know how many simple closed #geodesics are on your
hyperbolic Riemann surface?
• @MJMchem: There’s a hole in the ozone over Antarctica. Ban chlorofluorocarbons!!
• @nikola: I’m pretty certain my “alternating current” invention will help power cars
one day
• @PeanutExpert: We can fix our soil’s #nitrogen if we rotate crops today!
• @PhysicsAndChem: #polonium #radium
• @SurelyJoking: Oh hey, cool – you can do particle physics with diagrams!The solutions to the clues—many of whom were drawn from the list of innovators students
submitted on Monday—were as follows:
• @AstroMath: OMG I just figured out an incredibly efficient way to stack oranges!
I’m pretty sure it’s optimal!!! = Johannes KEPLER
• @BetaExpert: #ParityIsNotConserved = Chien-Shiung WU
• @BigHairPhys: My new theory is so #massive it might distort space-time! = Albert
EINSTEIN
• @computer: The trick to getting to and from space safely is parabola math! =
Katherine JOHNSON
• @dyNamite: Planning on endowing a prize for “benefit to humankind.” RT with
category suggestions, please = Alfred NOBEL
• @grace: So glad I was early enough to the computer revolution to be able to get my
first name as my username! = Grace HOPPER
• @HavanaEpi: Watch out for mosquitoes! They carry #yellowfever = Carlos FIN-
LAY
√
q p
• @HypergeometricMath: I just computed 1 + 2 1 + 3 1 + · · ·. = Srinivasa RA-
MANUJAN
• @MetaModuli: Want to know how many simple closed #geodesics are on your
hyperbolic Riemann surface? = Maryam MIRZAKHANI
• @MJMchem: There’s a hole in the ozone over Antarctica. Ban chlorofluorocar-
bons!! = Mario MOLINA• @nikola: I’m pretty certain my “alternating current” invention will help power cars
one day = Nikola TESLA
• @PeanutExpert: We can fix our soil’s #nitrogen if we rotate crops today! = George
Washington CARVER
• @PhysicsAndChem: #polonium #radium = Marie CURIE
• @SurelyJoking: Oh hey, cool – you can do particle physics with diagrams! =
Richard FEYNMAN
With those answers, the grid could be filled in as follows:
And as Becca Barbera, Maria Bayder, Omer Eyal, Lazar Ilic, Joonwoo Park, and Ellen
Xu figured out, reading down the highlighted column revealed “the identity of one additional
scientist whom I find particularly inspiring:” IT’S YOU, OF COURSE. (Thus, the correct
solution was to send in your own name as the answer.)
Thanks to everyone who solved this year’s Conundrums, and congratulations on a fan-
tastic ISEF! QED.You can also read
