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The emoji problem (2022)

by mtsolitary on 5/20/25, 10:18 AM with 62 comments

• by amenghra on 5/20/25, 1:24 PM

  There is a great quora answer for this one: https://www.quora.com/How-do-you-find-the-positive-integer-s...

• by adzm on 5/20/25, 2:06 PM

  On a related tangent, when I was teaching my younger kids about math and helping with homework, I would often rewrite things as a formula, or rewrite the formula itself once they got to that point. However instead of things like x I would use things like fluffy cloud, star, etc. They thought it was annoying but it still kept them interested and have said they've done the same when helping their peers. It's easy to forget what it was like learning these abstractions, and it was important to show that x was nothing special, it could be a sun, it could be the phrase "total number of kittens"

• by lblume on 5/20/25, 6:53 PM

  I tried giving this to ChatGPT. (Just by uploading the image to the base OpenAI interface.) I expected to either (a) have the model already know the question and give the right answer, (b) hallucinate an answer or (c) refuse to engage with the problem at all.

  Instead, this happened:

  https://chatgpt.com/share/682cce62-c53c-8003-be2c-2929395868...

  Basically, the model confidently outputs a guess, then calculates it, determines it to be incorrect, and repeatedly tries again, even repeating the same guesses over and over. It does not recognize any symmetry and acts like a completely unstructured agent. In the end, the model vehemently asserts there to be no solutions to this puzzle. I really did not expect this and will update my beliefs accordingly if the models behave as badly with future puzzles.

• by jsheard on 5/20/25, 1:51 PM

  > A guy by the name of Sridhar Ramesh

  Sridhar is a pro follow, you don't see many people with PhDs in both math and shitposting.

• by OisinMoran on 5/20/25, 12:11 PM

  I love this genre! I've started calling it “Dantzig Sniping” and here’s my own one: https://x.com/TheOisinMoran/status/1298305686082744320

  Some more context and related ones here: https://x.com/TheOisinMoran/status/1299124512240398336

• by oytis on 5/20/25, 11:05 AM

  It's year 2025, why doesn't the author use actual fruit emoji for variable names?

• by robinhouston on 5/20/25, 3:14 PM

  Using other constants in place of the ‘4’ can lead to some _really_ gigantic smallest solutions: https://observablehq.com/@robinhouston/a-remarkable-diophant...

• by ykonstant on 5/20/25, 10:52 AM

  I remember the problem when it first appeared, we had a good laugh at the number theory seminar :)

• by wredcoll on 5/21/25, 12:25 AM

  I appreciate the deep dive into the depths of number theory and strange graphs, but what is supposed to be tricky/confusing about the original apple/banana/etc puzzle?

  Is there supposed to be some easily confused bit so people argue or just be easy so everyone rushes to show off?

  (I got 10, 4, 2; maybe it confused me?)

• by zahlman on 5/20/25, 3:58 PM

  The "srsltid" in the query parameters is unnecessary: https://artofproblemsolving.com/community/c2532359h2760821_t...

• by less_less on 5/20/25, 11:57 AM

  The theory of elliptic curves goes amazingly deep. Scratching the surface slightly more, to fill in the article's "ignore the labels like 2P":

  Intersecting the curve with lines the way the author does is, perhaps shockingly, a commutative group operation, known as point addition. You define this operation by saying that the three points A,B,C on a line sum to zero: that is A+B+C=0 or in other words, A+B = -C. Reflecting across the curve's line of symmetry is negation (there's an alternative definition that extends to curves without reflection symmetry). Combining the two defines an operation A+B which adds two points on the curve and gives a third one: draw the line from A to B, intersect it with the curve to find a point -A-B (using the same type of formula given in this article) and then reflect it to get A+B. This addition operation obviously commutes (meaning, the line between A and B is the same as the line between B and A), but surprisingly it also associates and you get a group operation.

  (For math olympiad nerds out there: so the union of a conic and a line is also a bivariate cubic equation. You can carry out the same "addition" operation there. Again the operation clearly commutes. But it also associates! This is basically Pascal's theorem.)

  The theory of elliptic curves is also the basis of elliptic curve cryptography. In that case, instead of the curve being over the reals, all the calculations are done mod some prime p, which destroys structure based on continuity and prevents the numbers from becoming too large. There are a bunch of subtleties here but the key is that you can still straightforwardly compute addition in this context, with basically the same formulas. Then from addition you can get n*A, both for small integers n (eg 5*A = A+A+A+A+A) but also for large n (e.g. 2*A = A+A; 66*A = 2*2*2*2*2*2*A + 2*A). This "scalar multiplication" operation is a one-way operation for appropriately chosen curves: it's easy to calculate n*A from (n,A), but as far as we know it is hard to calculate n from (A,n*A) ... at least if you can't build a large quantum computer.

  This gives you mix of easy operations (eg, addition and scalar multiplication) plus problems believed to be hard, which is a great starting point to build cryptography. There are a lot of important technical details, though fewer than with the new lattice schemes.

  (Further comment on the olympiad thing: so you can do this with conic+line too, extending from addition to multiplication and using the same formulas mod p. But it's not as good: it's not hard to find a projection that sends the line to infinity and the conic to the unit circle or similar, and then the group operation becomes equivalent to multiplying the points' coordinates as complex numbers or similar. If the group operation is equivalent to multiplication, then scalarmul becomes equivalent to exponentiation, which ends up being in Fp or Fp^2 depending on some Legendre symbol or other. Exponentiation is still potentially secure: it's basically classical Diffie-Hellman. However, more attacks are known on exponentiation in Fp or Fp^2: the attacks don't outright break it but you need p to be much bigger.)

  Edit: unescaped stars make italics.

• by rolandhvar on 5/20/25, 5:49 PM

  Dumb question, but how do you know it's smallest?

• by gcanyon on 5/20/25, 11:04 AM

  Once upon a time I was good at math -- like, win state-wide competitions, get a scholarship good. It's sad that I can easily follow the path laid out here, but I am nowhere close to being able to lay out this course on my own.

• by raverbashing on 5/20/25, 11:10 AM

  So the meme was a crap-post and never "supposed" to be solved anyway (in human terms?)