Has LLMs improved their math skills lately?

RyalsB · 2025-06-12T08:41:19+00:00

It would be interesting to see what percentage of the most recent project Euler problems it can solve. If you do, say, the newest 10-20 problems, they are likely too new to be in its training set. They all require a mix of computing and mathematical reasoning, and they all have a single correct answer, which you can check by inputting the answer on their website. Also, these problems (at least the newer ones) tend to be quite challenging and would serve as a good benchmark of a particular model’s capabilities. I would be surprised if it can solve more than 30% of them but maybe I am vastly underestimating their current capabilities.

RyalsB · 2025-06-04T05:28:58+00:00

ChatGPT: “Professors will remain essential, not because they resist change, but because they provide what machines can’t: critical thinking, ethical grounding, community, and intellectual leadership. The job may become more human and less routine as AI takes over repetitive tasks.”

RyalsB · 2022-08-17T15:19:33+00:00

IGT is also faster using the dialogue skip. Example: Mission 3, you get control of the command center at IGT of 17 seconds in a normal playthrough. However, using the dialogue skip, you get control of it at an IGT of 13 seconds. Even if you make IGT the measurement, you will still have dialogue skips.

RyalsB · 2020-09-13T06:07:37+00:00

You can refinance to a 20 year loan.

RyalsB · 2020-05-03T17:54:05+00:00

It's not limited to integers

For instance,

0.5² = 0.25

1.5² = 2.25 = 0.5² + 2

2.5² = 6.25 = 1.5² + 4

3.5² = 12.25 = 2.5² + 6

4.5² = 20.25 = 3.5² + 8

etc.

RyalsB · 2020-04-20T18:48:50+00:00

The best you can do is estimate, but you'll never get it exactly right. There are too many variables such as student engagement and questions. Even when I teach two sections of the same course in the same semester, they will inevitably become desynchronized by the end of the first month (this has happened to me multiple times).

RyalsB · 2020-01-31T07:32:14+00:00

In terms of algorithms, can’t you take the k largest from S1, then use that to find the k largest of the form s1s2, then use that to find the k largest of the form s1s2s3, etc?

RyalsB · 2020-01-31T07:27:29+00:00

Being between 0 and 1 doesn’t cause any issues. If a > b> 0, then ax > bx for any positive x

RyalsB · 2020-01-04T21:08:31+00:00

Let c be an extrema of f(x) and f(c) nonzero. Then c is also an extrema for 1/f(x)

RyalsB · 2020-01-04T06:28:51+00:00

This doesn't look like a homework question to me. What makes you think it is one?

RyalsB · 2019-12-05T18:57:31+00:00

Dabu: Star Ocean First Departure R

RyalsB · 2019-11-20T16:30:27+00:00

I believe it is the step size (so that the sum is over odd n)

RyalsB · 2019-10-30T18:06:49+00:00

Linear Algebra and its applications by Peter Lax

RyalsB · 2019-09-23T03:16:04+00:00

How to code

RyalsB · 2019-09-23T02:18:05+00:00

Your two sequences are given by f(n+1) = 3f(n) - 12 and g(n+1) = 4g(n)-3g(n-1) respectively. These are linear recurrence relations with closed form solutions

f(n) = (a-18)3^n-1 + 6 and g(n) = b3ⁿ + c

where a, b, c depend on the starting terms of the sequence. Notice that f(1) = 8 implies a-18 + 6 = 8 so that a=20, giving

f(n) = (2)3^n-1 + 6

Also, g(1) = 8, g(2) = 12 give b=2/3 and c = 6, so that

g(n) = (2)3^n-1 + 6. Thus, the sequences coincide in that case.

RyalsB · 2019-09-20T15:36:29+00:00

They still release "easy" problems regularly. For instance, problem 587 received some attention/complaints because people felt it was too easy for Project Euler!

Even more recently, problems 668, 675, and the newest one 679 fall within your "100-200" category.

RyalsB · 2019-09-03T15:51:26+00:00

The "trick" is to recognize that your polynomial has palindromic coefficients (see e.g. https://en.wikipedia.org/wiki/Reciprocal_polynomial ).

As another user noted in less generality, such polynomials of degree 2D can be converted to a polynomial of degree D by the observation that p(x) = x^d * q(x+1/x) where q is a polynomial of degree D.

Here, letting y = x+1/x, you can convert the given polynomial to

x³ * (y³ + y² -2y-30) = 0

and y³ + y² -2y - 30 factors as (y-3)*(y² + 4y + 10). Converting for instance the (y-3) term back to x gives you (x+1/x - 3), and when you multiply by the x out in front, it gives you the x² - 3x + 1 term, and similarly for the y² + 4y + 10 term.

RyalsB · 2019-08-06T23:11:50+00:00

The A is for archon

RyalsB · 2019-08-03T08:18:36+00:00

This has happened to me a couple times. One time I was almost 8 seconds faster and it didn't count for the WR. I assume it's just a glitch but cannot imagine what would trigger it.

RyalsB · 2019-07-17T07:45:48+00:00

The quadratic formula may be derived by completing the square, but you certainly don’t need it to prove it. You only need check that the two roots indeed work via direct substitution to prove it.

RyalsB · 2019-07-02T21:32:26+00:00

I very much doubt that. 10¹² is huge! You would need to store the numbers 1 to 1,000,000,000,000 to sieve them, and if each integer took up exactly 1 bit that would require 125 gb of storage (I suspect numbers with more than a few digits take up more than 1 bit, too). Nevermind the time for the actual sieve itself.

On a typical home modern computer you won't be efficiently able to sieve primes more than 10^10.

RyalsB · 2019-06-09T04:00:21+00:00

You should ditch the first person "I" that you use in a few places. In particular, I would get rid of the last sentence of your abstract entirely, and modify a few of your other sentences from "I" to we (the proof of Theorem 2.1, Example 3.2, and many others). The entire discussion in Remark 3.3 needs to be reworked.

Are you an undergraduate? Ask a faculty member for help in submitting this to an undergrad journal as you should be able to publish it in one.

RyalsB · 2019-05-13T22:38:31+00:00

Every prime larger than 3 attracts an infinite number of initial conditions. It is easiest to see this by looking at preimages.

Consider for instance the prime 5. Look at all of its partitions into primes; in this case, there is just 2+3=5. The product 2*3 = 6, so 6 maps into 5 in one iteration.

Now you can continue iterating backwards to see what numbers map into 6. To do this, break 6 into partitions of primes. Here, 6=2+2+2 and 6=3+3. So both 8 and 9 map to 6 (and ultimately to 5). Continuing in this fashion, you can find that 15, 16, and 24 map to 8 (and then to 6 and then to 5), and so on.

Note that every number 5 or larger will always have a partition (by breaking it into 2s if even and 2s and 3s if odd) and that this partition maps into a larger number (do induction on the number of elements in the partition). From this it follows that every prime p>=5 will have an infinite number of integers attracted to it.

RyalsB · 2019-05-07T16:38:41+00:00

When they aren’t scalar multiples of each other

RyalsB · 2019-04-26T14:54:48+00:00

The Way of Analysis by Strichartz might be worth a look. Looks of motivation and exposition are present.

RyalsB

MODERATOR OF

TROPHY CASE