# How would you have answered Richard Feynman's challenge?

Reading the autobiography of Richard Feynman, I struck upon the following paragraphs, in which Feynman recall when, as a student of the Princeton physics department, he used to challenge the students of the math department.

I challenged them: "I bet there isn't a single theorem that you can tell me ­­what the assumptions are and what the theorem is in terms I can understand ­­where I can't tell you right away whether it's true or false."
It often went like this: They would explain to me, "You've got an orange, OK? Now you cut the orange into a finite number of pieces, put it back together, and it's as big as the sun. True or false?"
"No holes?"
"No holes."
"Impossible! There ain't no such a thing."
"Ha! We got him! Everybody gather around! It's So­-and­-so's theorem of immeasurable measure!"
Just when they think they've got me, I remind them, "But you said an orange! You can't cut the orange peel any thinner than the atoms."
"But we have the condition of continuity: We can keep on cutting!"
"No, you said an orange, so I assumed that you meant a real orange."
So I always won. If I guessed it right, great. If I guessed it wrong, there was always something I could find in their simplification that they left out.

Forgetting that most of this was probably done as a joke, with what theorem would you have answered to Feynman's challenge?

• Almost anything non-trivial from Combinatorics (assuming the statement requires no simplifications). Dec 5 '21 at 17:58
• (a) 0 is not the sum of three non-zero integer cubes. (b) But 33 is. Dec 5 '21 at 18:45
• OK, if we are limiting to results before 1988: aperiodic tilings exist. en.wikipedia.org/wiki/Aperiodic_tiling Dec 5 '21 at 19:00
• Given Feyman's prowess in solving differential equations, Lewy's example might be another candidate. en.wikipedia.org/wiki/Lewy%27s_example Dec 5 '21 at 19:35
• Sphere eversion seems like a reasonable candidate to me. Dec 5 '21 at 20:19

There's a certain gaming/sporting aspect to Feynman's challenge that works in his favor. First of all, as phrased, the challenge gives him a 50/50 shot at being right even if he guesses randomly. Also, if you present a statement which seems too obviously true then Feynman could reason that you wouldn't have chosen that statement if the obvious answer were correct.

With those caveats, I present some proposals below. I have tried to gravitate toward problems involving physical intuition, since IMO fooling Feynman's physical intuition carries bonus points.

1. Is sphere eversion (without creasing) possible? (Edit: I see now that Noah Schweber already suggested this one.) This is my favorite, and the only flaw is that a real physical surface can't pass through itself, so the question isn't quite a physical one.

2. Is there a convex solid of uniform density with exactly one stable equilibrium and one unstable equilibrium? This problem postdates Feynman's life and maybe he would have guessed correctly, but I would be impressed.

3. Does the regular $$n$$-gon have the largest area among all $$n$$-gons with unit diameter? Tricky because the answer is yes if $$n$$ is odd but false if $$n$$ is even and $$n\ge 6$$.

4. Is there a closed planar convex shape with two equichordal points? This one is good if you suspect that Feynman might reason that the "obvious" answer (no) can't be correct or else you wouldn't be asking. "No" is correct but the proof is highly nontrivial.

5. Does every $$d$$-dimensional polytope have a realization in which all its vertices have rational coordinates? Explaining the precise statement of this result is a little tricky, and it has the flaw that the weirdness doesn't kick in until $$d=4$$, but otherwise this one is really nice IMO.

EDIT: There has been some discussion in the comments about whether the presence of irrational numbers in #5 above means it refers to some mathematical abstraction that is not "physically realizable." Here is an easier-to-understand (though less spectacular) question that captures the essential issue. Consider the Perles configuration of nine green points in the figure below. If the outer pentagon is a mathematically perfect regular pentagon, then at least some of the nine points have to have irrational coordinates. No big surprise there.

But now consider this. Suppose I don't require that the pentagon be a perfectly regular pentagon; suppose I allow you to redraw the figure, re-positioning the nine points in any way you like, as long as you preserve all the collinearities (i.e., points that are collinear in the original diagram must remain collinear in your new picture). Can you arrange for all nine points to have rational coordinates? The answer, which I find surprising, is no.

As far as physical intuition is concerned, what this example shows is that the seemingly physical concept of "collinearity of points in the plane" (suitably idealized of course) already forces irrational numbers upon us. In contrast, if you point out that a right-angled triangle with two sides of length 1 has a hypotenuse of $$\sqrt{2}$$, then a physicist could object that physical angles and lengths are never infinitely precise, and that you can create an arbitrarily close approximation using rational numbers. In the Perles configuration, we are not stipulating any distances or angles precisely; we are only stipulating that certain points be exactly collinear, and it turns out that this idealization already requires us to introduce irrational numbers.

• Reading some of the other answers, I propose the following criterion for a good challenge: The wrong answer looks significantly more plausible than the right answer. So challenges that clearly are random coin flips unless you do a lot of calculation (is the billionth digit of $\pi$ even or odd?) are not very good. Dec 6 '21 at 4:02
• Great answer! The question doesn't say that, but my sense is that Feynman may have had in mind more geometrical, topological areas of math, rather than number theory and such. Dec 6 '21 at 5:39
• @ManfredWeis Spoken like a true Platonist! A fictionalist would say that there is no such thing as a mathematical surface; only physical surfaces actually exist, and a mathematical surface is a fictional idealization. Dec 6 '21 at 17:33
• The biggest little polygon problem is a great example, but an example like #5 is in a different category because it depends on the distinction between rational and irrational numbers. Feynman's example about the orange suggests that he has what is to a physicist a natural aversion to questions like this that sound like they're about a physically realizable surface, but in fact can't be realized by any possible physical process.
– user21349
Dec 6 '21 at 18:25
• @TimothyChow: I like your "wrong answer vs. right answer" criterion. Speaking as a physicist, the "biggest little polygon" example is particularly good because it leads me into the physicist's trap of "the symmetric answer is the most plausible answer". Dec 7 '21 at 19:04

Can you hear the shape of a drum?

Easy to understand problem for a physicist, with a non-trivial answer given by Gordon, Webb and Wolpert in 1990.

• I think this is the best. It is basically a physics question with a non trivial solution. Although an affirmative answer would have been more surprising (to me). Dec 6 '21 at 10:06
• @lalala Right. I thought of this one too but rejected it since I imagined Feynman would have immediately guessed that the answer is no. Dec 6 '21 at 11:50
• A possibly better question: Can you hear the difference between a square and a circular drum? (The answer to this one is yes) Dec 6 '21 at 14:47
• This is too easy if you are gaming the questions. Immediately answer no. If the answer is no, you are right! (and it turns out that you are right in the general case) If the answer is yes, listen for a while to the explanation and then say "But you said hear the shape! So I assumed you meant hear with my ears, and my sense of pitch is not that good." It's basically the same as "I assumed you meant an actual orange". Dec 6 '21 at 20:20
• You could also game the question the other way. Immediately answer yes. When you are told that the answer is no, listen for a while to the explanation and say "But these shapes are very complicated. When you said a drum, I assumed you meant a real drum. You can't make the boundary conditions of the drum finer than the atoms, and then I bet you can't realize it." Actually, is there a form of theorem where the boundary can only be realized to some precision and you want the sound of the drum to be statistically indistinguishable? Dec 6 '21 at 20:56

There are six Platonic solids in 4 dimensions.

If you have never thought about this problem before, I think it's almost impossible to deduce the truth of this statement immediately, as Feynman sought to do.

Here are two questions, and both are about math that was known long before Feynman passed away.

1. Explain to him what unique factorization into irreducibles means (including the ambiguity from multiplication by invertible elements, like the nonzero constants in polynomials over $$\mathbf R$$) and what a "quadratic number system" $$\mathbf Z[\sqrt{n}]$$ is, where $$n$$ is not a perfect square. Tell him $$\mathbf Z[\sqrt{n}]$$ has unique factorization into irreducibles when $$n = 2$$ and 3, but not when $$n = 5$$. Now we're ready for the questions: ask him one after the other about whether or not there is unique factorization in $$\mathbf Z[\sqrt{n}]$$ for $$n = 6$$ (yes), then for $$n = 7$$ (yes), then for $$n = 8$$ (no), then for $$n = 10$$ (no), and so on. It's the same question for each $$n$$, so there's no issue of him not understanding the question after he deals with the first few. Note the question for each $$n$$ has a definite yes/no answer, even though it is still an open problem to show infinitely often (for $$n > 0$$) there is unique factorization. If $$\mathbf Z[\sqrt{n}]$$ is not the full ring of integers of $$\mathbf Q(\sqrt{n})$$ then it does not have unique factorization, and if it is the full ring of integers of $$\mathbf Q(\sqrt{n})$$ then a table of class numbers will indicate whether or not $$\mathbf Z[\sqrt{n}]$$ has unique factorization. There is no mathematical or physical intuition that can help someone answer this question correctly right away for each $$n$$, so he'd eventually be wrong and not be able to "explain" his wrong answer. Let's keep in mind that his constraint was that the problem had to be expressed in terms he could understand, not that it has to be about a part of math that appeals to him (this addresses a point made in the comments above).

2. If $$n > 1$$ is not a perfect square, does each equation $$x^2 - ny^2 = 1$$ have a solution in nonzero integers $$x$$ and $$y$$? Depending on what he says to that, maybe ask him about $$x^3 - ny^3 = 1$$ always having a solution in nonzero integers when $$n$$ is not a perfect cube. Or change to "infinitely many integral solutions" for those two kinds of equations.

• Ah, yeah, the Pell's equation (units theorem for real quadratic fields) is an excellent example, a first instance of the irregularity of number-theoretic things. :) Dec 5 '21 at 19:29
• Although this response is technically correct, I feel that it is not quite in the spirit of Feynman's challenge, because morally speaking, you're presenting a parametrized family of theorems rather than a single theorem. The challenge is "too easy" if we're allowed to ask Feynman to compute $f(n)$ for $n=1,2,3,\ldots$ for an arbitrary hard (Boolean) function $f$ of our choosing. Dec 5 '21 at 21:20
• @TimothyChow then you could ask him a specific case: does $\mathbf Z[\sqrt{79}]$ have unique factorization? (answer: no, the class number is 3.) Or pick a Mordell equation: is there a perfect square and perfect cube that differ by 1989? (No, but I admit I don't know when that was settled.) By 1992? (yes: $y^2 - x^3 = 1992$ for $(x,y) = (14833, 1806523)$. Dec 5 '21 at 21:56

Exactly one of the numbers, $$2^{11213}-1,2^{11239}-1$$ is prime. Which one?

• What is the answer, or where should I look for an answer :)
– Amr
Dec 6 '21 at 8:32
• @Amr, if you got any mail from the Math Dept at U Illinois in the late 1960s, you would have seen it was postmarked with "$2^{11213}-1$ is prime". Dec 6 '21 at 8:35
• Not a true or false question. "I bet there isn't a single theorem that you can tell me ­­what the assumptions are and what the theorem is in terms I can understand ­­where I can't tell you right away whether it's true or false." Dec 6 '21 at 15:13
• @DougSpoonwood It's trivial to turn this into a true/false question. Assumption: exactly one of $2^{11213}- 1$ and $2^{11239} - 1$ are prime. True or false: $2^{11213} - 1$ is prime and $2^{11239} - 1$ is not. Dec 6 '21 at 17:10
• @NuclearHoagie Why exactly did you ask if the first number was prime instead of the second? Seems like it implies two true or false questions equally so to me. Dec 6 '21 at 20:03

Every positive integer is a sum of fourth powers: trivially, $$1+1+\dotsb+1$$. Of course we can use fewer summands. For example, $$79$$ is a sum of $$4$$ $$16$$’s and $$15$$ $$1$$’s, needing $$19$$ summands. Is there any positive integer that needs more than $$19$$? The answer is no, shown in 1986 by Balasubramanian et al.

• Oh, forgot to be tricky. Okay, 15 requires 15 summands, 31 requires 16, clearly it goes up more, but does it get past, say, 20? Dec 5 '21 at 23:47

Assuming that Feynman understands integers and limits:

Let $$S$$ be a set of positive integers such that $$S, 2S, 3S$$ are disjoint. Then the upper density

$$\limsup_{N \to \infty} \frac{\# \{ S \cap [1, N]\}}{N}$$

of $$S$$ is at most $$\frac{108735}{117936}$$, and this bound is sharp.

True or false?

• I guess Feynman (Nobel Prize in Physics 1965) understood integers and limits. See in particular "Feynman path integral". Dec 6 '21 at 4:35
• Is it true or false? (Maybe Feynman would have known, but I don't. If you would have asked me to guess, I would have guessed 6/11.) Dec 6 '21 at 14:30
• @GHfromMO Considering how quickly any naive formulation of the path integral gets you to something physically useful yet mathematically ill posed, I am not sure if knowing too much about limits might not actually be detrimental in understanding it...
– mlk
Dec 6 '21 at 14:32
• @Will Brian It is false, but the actual answer is very close to what’s written. I will ask the person who sent me the result whether he is okay with me sharing his answer and proof of the result! Dec 6 '21 at 21:19
• Yes, the comment was tongue in cheek haha Dec 7 '21 at 2:57

Imagine a brick of weight $$1$$ trapped between a wall and a brick of weight $$100^n$$, coming at it with some speed. Assume no friction and elastic collisions. True or false: the number of collisions between the bricks and the smaller brick with the wall is the sequence of the first $$n+1$$ digits of $$\pi$$.

• This video is absolutely amazing. Dec 10 '21 at 17:54
• @Nucleon, I love 3b1b videos; he does an excellent job explaining things simply and clearly, and he chooses the most beautiful math connections for his videos. Dec 15 '21 at 22:36
• This is still open right? Dec 21 '21 at 11:40
• @VilleSalo, the video goes over the solution. 3b1b has quite a few very beautiful videos about unusual results. The core of this solution is to trace the movements of the bricks in the configuration space, the axis corresponding to the bricks position; not too difficult in the hindsight, when you know the result, but completely unexpected beforehand. Dec 21 '21 at 16:33
• I think the solution assumes that $\pi$'s decimal expansion doesn't have massive conspiracies though. Dec 21 '21 at 17:09

In terms of a math problem that a lay person* could understand but have completely the wrong intuition about, rather than anything involving deep mathematics, I might choose something from elementary probability: e.g., the Monty Hall Problem or Simpson's Paradox. As the Wikipedia link says, even Erdös gave the wrong answer to the Monty Hall Problem. And Simpson's Paradox seems at first blush to contradict basic probabilistic facts like linearity of expectation. These are just two examples; the point is that our brains seem to have a lot of (not always correct) probabilistic intuition hard-wired into them.

*Which of course Feynman was not, though he was also not a mathematician.

• I considered this kind of thing but decided that it wasn't in the spirit of Feynman's challenge to state a theorem, rather than a puzzle or a problem. But if we allow it, I'd nominate the puzzle in Gil Kalai's 30th Test Your Intuition. You throw a normal 6-sided die until you get 6. What is the expected number of throws (including the throw giving 6), conditioned on the event that all throws gave even numbers? Dec 6 '21 at 3:45
• The Monty Hall problem is rather difficult to state precisely. A lot of statements of it (perhaps the majority) would allow Feynman to say either "yes, you should switch" or "no, there's no advantage to switching," at his whim, and then explain the flaw in that statement that allowed him to answer it that way. Dec 6 '21 at 10:41
• In particular, a lot of statements of the problem merely describe a sequence of events and fail to explain what Monty's general behavior is. In such statements, the correct answer could be anything from "switching is a guaranteed win" to "switching is a guaranteed loss." Dec 6 '21 at 10:43
• It's much easier to understand the Monty Hall result if you regard the guest as choosing one of two doors that does not hide the car, the host choosing the other. But even with that in mind, it doesn't help my intuition when faced with the original statement of the problem. Dec 6 '21 at 19:27

Equip the sphere $$\mathbb{S}^2$$ with any Riemannian metric, then it must have at least three closed geodesics by the theorem of three geodesics. I believe these are terms which Feynman should understand as he has to know General Relativity.

As for the Banach Tarski paradox, I think the mathematicicans really got him but he doesn't want to admit it :D His response about a real oragne means that he is really challenging the mathematicians to tell him a physical fact he didn't know and not a mathematical fact he didn't know :)

• If I were asked how many closed geodesics a Riemannian metric on the sphere has, at minimum, I would guess the ellipsoid, the only surface for which I can immediate guess the geodesics and for which the number is smaller than for the sphere, must give the right answer. I think Feynman would get that one right. But you might wonder if the answer is the same for Riemannian metric and for Finsler metrics, and then I think he would probably get it wrong. Dec 6 '21 at 17:38
• @BenMcKay I agree.
– Amr
Dec 7 '21 at 3:24

Claim 1: It is possible to write a program which decides whether any other computer program will ever finish running. (This is the halting problem).

It can't. Can be proved using a diagonal argument.

By the Riemann integral, I mean the integral for all continuous functions from $$[0,1]$$ to $$\mathbb R$$, which we can prove always exists.

Claim 2: The Riemann integral is computable.

Claim 3: Finding an eigenvector of a positive-definite symmetric matrix is computable.

In conclusion:

See how many of those you got right.

• Although properly explaining what computability in the present context means, I think those make for some fine examples. A simpler and, I think, an even more surprising example is the fact that finding derivatives (or even determining differentiability in the first place) is not a computable problem. Dec 6 '21 at 14:42
• @Wojowu This is subjective but I find the uncomputability of differentiation not too impressive, as differentiation is defined by a limit, and one shouldn't expect a limit to be computable. The halting problem provides some intuition that "infinitary" problems are not solvable by computers. The Riemann integral example shows that some "infinitary" problems are solvable by computers, contradicting one's intuition. That doesn't make such a positive computability result practical though.
Dec 6 '21 at 14:57
• "...and one shouldn't expect a limit to be computable" Well, I certainly would have expected limits to be computable, had I not known any better! Dec 6 '21 at 15:16
• @Wojowu Let $T$ be a Turing machine. Let $u_n(T) = \begin{cases}1, &\text{if$T$has terminated before the$n$th step} \\ 0, & \text{otherwise} \end{cases}$. The limit $\lim_{n \to \infty} u_n(T)$ always exists, but is not computable in general on pain of solving the halting problem. I think upon realising this, most people would reflexively conclude that integration is not computable along with differentiation, as those are a very general class of limits.
Dec 6 '21 at 15:26
• @WorldSEnder This would do it. link.springer.com/book/10.1007/978-3-642-56999-9 . There are no additional assumptions. The functions which you're integrating are represented as programs on a Type Two Turing Machine (the subject of the book I've linked to). Those functions are all automatically continuous. The integral of a continuous function $f : [0,1] \to \mathbb R$ always exists, as I pointed out
Dec 6 '21 at 18:21

Can every solvable position of the 2x2x2 Rubik's Cube be solved in fewer than 12 moves?

Answer: Yes, first computed in 1981.

1. "Your job is coloring maps. You can't color two countries which touch each other along an edge the same color, but if they touch at a point it's ok if they are the same color. You have only $$N$$ different colors, can you color every possible map?" Ask this question for various $$N$$ to see if they can figure out $$N=4$$ suffices. Now, the kicker: "So, someone comes to you with a map drawn on the surface of a torus. How many colors do you need?" Actually I think the answer to the second question was proven before the answer to the first was (and both dates are later than the story, alas), but I bet a lot more people know the answer to the first.

2. I thought I had a great question, but then I realized how to wiggle out. The question was something like "You are going on a spy mission. At some point, after collecting all the information, you'll radio it back to us. You'll get a lot of information, and it will take $$10^6$$ letters to write it all down. Of course, in advance we will agree on some code so that the enemy can't learn what you say. Unfortunately, you won't be able to hear anything from us, and because of noise, for each letter you send there is only a 50% chance we get that letter, independent of the chance we get any other letter. So, roughly how many letters do you need to send to ensure that we have at least a 99% chance of getting your whole message perfectly?" This is being a bit bold as a question, because the answer involves entropy and a physicist might be able to get a reasonable guess at it, but statistical physicists do better at this kind of question than high energy, so I'll go with it. But, the wiggle out is that potentially after explaining the solution, one can just say "but I don't have any way of performing those calculations! A computer capable of doing that would take up an entire room! And you said I was a spy!" After all, this was a long time ago...

The Continuum Hypothesis. Make it into either a positive or negative assertion and let him decide whether it is true or false!

I guess with combinatorics and number theory it's too easy...

Topology/geometry:

Every map $$S^3\to S^2$$ is homotopic to a constant. True/False? (But probably Feynman knew about the Hopf map)

There are orientable 3-manifolds which are not boundaries of compact 4-manifolds. True/False? Same question for 4-manifolds.

Two differentiable.structures on $$\mathbb R^4$$ are diffeomorphic. True/False? Same question for $$\mathbb R^5$$.

Analysis:

The partial sums of the Fourier series of a continuous function on the circle converge uniformly. True/False? Same question for $$L^1$$ functions, with convergence in $$L^1$$.

There exist $$C^1$$ isometric embeddings $$S^2\hookrightarrow\mathbb R^3$$ whose image has diameter less than $$\pi$$. True/False?

An n-dimensional function is differentiable in a point x, iff the directional derivative in x is linear in the directions, and if the chain rule holds in x for differentiable curves through x.

1. The Frobenius Conjecture:

Let $$G$$ be a finite group, let $$n$$ be a divisor of the order of $$G$$, and assume that the set $$H$$ of $$n$$th roots of unity in $$G$$ has size $$n$$. Then $$H$$ is a subgroup of $$G$$.

I think Feynman would have understood the assumption and the statement of this theorem, but its proof uses the Classification of Finite Simple Groups, and I think it isn’t intuitively obvious, at least not to me, that $$H$$ should be a subgroup or why it should be a subgroup — it seems almost a “coincidence”.

1. Tutte's Planar Embedding Theorem, rephrased as follows:

Let $$G$$ be a $$3$$-connected planar graph, and let $$F$$ be a face of $$G$$. Realize $$G$$ as rubber balls (vertices) connected by springs (edges) with spring-constant $$1$$. Place $$G$$ on a wooden surface, and nail the vertices in $$F$$ to the wood to form a circle, large in relation to $$G$$ (say, the radius is more than all the balls placed in a long line). Assume negligible friction between the wood surface and the balls. Under the force of the springs, $$G$$ will settle into a planar embedding.

• For 1, I'm assuming "$n$th roots of unity" refers to $n$-torsion points. In the statement one should also assume $|H|=n$ (otherwise elements of order $2$ in $S_3$ form a counterexample) Dec 7 '21 at 10:02
• @Wojowu Solutions to $x^n = 1$. Dec 7 '21 at 10:58
• The formulation of Frobenius conjecture here is quite different from the formulation in the Wikipedia article. As far as I can tell, the statement in the answer is trivial: the set of $n$th roots of unity is definable, hence invariant under all automorphisms (actually, even endomorphisms); i.e., the subgroup of $G$ they generate is fully characteristic, and a fortiori normal. In particular, if $n$th roots of unity form a group by themselves, this group is automatically normal. Dec 7 '21 at 13:07
• I took the liberty to fix the statement. Dec 7 '21 at 13:15

The Hadamard and de la Vallee-Poussin theorem that $$\zeta(s)$$ does not vanish on the line $$\Re(s)=1$$.

(Of course, in real-life terms, such games there would have been dangerous... I guess RF did not see anyone among the grad students who could out-guess him, at that particular time. In a way, bad luck for him.)

I have read that Feynman had trouble believing that the P vs. NP problem was even an open conjecture.

So perhaps a meta-question that might be hard for Feynman to wiggle out of would be along the lines of

• True or False: there is a problem that is known to be in NP but known not to be in P, or
• There is a problem that is known to be in NP but known not to be in CoNP, or even simpler
• It is known that TSP cannot be solved in polynomial time.

The regular tetrahedron is the only compact surface in $$\mathbb R^3$$ that has arbitrarily long closed geodesics.

A lot of cryptography is based on posing problems where the answer is known and easy to check, but extremely hard to find. Moreover, the problems are often straightforward questions in arithmetic. Thus, these kinds of problems seem ideal if you want to win such challenges.

As an example, here is an RSA Type Challenge.

Choose an $$n$$ according to the current computing power available. Choose two $$n$$-bit prime numbers $$p$$ and $$q$$, and calculate $$n=pq$$. Also choose a smallish prime $$r$$ (we take $$r=17$$ below) and choose a random number $$s$$ between $$1$$ and $$r-1$$. The question you pose is: "Is the remainder of the smaller of the two factors of $$n$$ modulo 17 equal to $$s$$?"

• The probability of guessing the right answer is $$1/16$$ which is small enough.
• It is easy to generate a few more of these problems, or to raise the size of $$r$$ in case you are worried about a correct guess!

Having said that:

• RSA was itself supposedly a secret until the 70's, so perhaps no one could/would have asked this specific question! (One can pose similar Diffie-Hellman challenges.)
• It seems like cheating to pose a question which you know that you would also not have been able to solve had you not known the answer!
• Note that the answer (to the above statement) could be easier to find than the factors of $$n$$. However, I do not know of an algorithm that is simpler than factoring.
• The way the challenge is posed, it must be a "true"/"false" question, so the probability of guessing correctly must always be 0.5 if Feynman guesses uniformly randomly. Dec 8 '21 at 18:30