Figure 1.
Logo of the International Mathematical Olympiad
. Source.
A waste of time like the rest of the knowledge olympiads.
To be fair, this is one of the least worse ones.
Their system is quite good actually. Not as good as a GitHub repo with all the tests made explicit. But still pretty good.
They don't have an actual online judge system, all problems simply have an integer or floating point solution and they just check that you've found the value.
The only metric that matters is who solved the problem first after publication, e.g.: projecteuler.net/fastest=454. The "language" in which problems were solved is just whatever the user put in their profile, they can't actually confirm that.
Project Euler problems typically involve finding or proving and then using a lemma that makes computation of the solution feasible without brute force. As such, they live in the intersection of mathematics and computer science.
Code solutions by individuals:
Basically no one ever had the patience to solve them all. What we need is a collaborative solution.
Once you solve a problem, you can then access its "private" forum thread: projecteuler.net/thread=950 and people will post a bunch of code solutions in there.
projecteuler.net says it started as a subsection in mathschallenge.net, and in 2006 moved to its own domain. WhoisXMLAPI WHOIS history says it was registered by domainmonster.com but details are anonymous. TODO: sample problem on mathschallenge.net on Wayback Machine? Likely wouldn't reveal much anyways though as there is no attribution to problem authors on that site.
www.hackerrank.com/contests/projecteuler/challenges holds challenges with an actual judge and sometimes multiple test cases so just printing the final solution number is not enough.
The beauty of Project Euler is that it would serve both as a AI code generation benchmark and as an AI Math benchmark!
TODO: real name? Occupation?
Claude says he's from the UK and has a background in mathematics. Oxbridge feels likely. How I Failed, Failed, and Finally Succeeded at Learning How to Code says he started off on the ORIC computer, which is British-made, so he is likely British.
This was a registration CAPTCHA problem as of 2025:
Among the first 510 thousand square numbers, what is the sum of all the odd squares?
Python solution:
s = 0
for i in range(1, 510001, 2):
    s += i*i
print(s)
At: euler/0.py
Solution:
233168
Solutions to the ProjectEuler+ version:
The original can be found with:
printf '1\n1000\n' | euler/1.py
A(x) = x + 1
Z(u)(v) = v
S(u)(v)(w) = v(u(v)(w))
Let's resolve the second example ourselves:
S
  (S)
  (S(S))
  (S(Z))
(A)
(0)

S
(S)
(
  S
  (S(S))
  (S(Z))
)
(A)
(0)

S
(S(S))
(S(Z))
(
  S
  (
    S
    (S(S))
    (S(Z))
  )
  (A)
)
(0)

S
(Z)
(
  S(S)
  (S(Z))
  (
    S
    (
      S
      (S(S))
      (S(Z))
    )
    (A)
  )
)
(0)

S(S)
(S(Z))
(
  S
  (
    S
    (S(S))
    (S(Z))
  )
  (A)
)
(
  Z
  (
    S(S)
    (S(Z))
    (
      S
      (
        S
        (S(S))
        (S(Z))
      )
      (A)
    )
  )
  (0)
)

S
(S)
(S(Z))
(
  S
  (
    S
    (S(S))
    (S(Z))
  )
  (A)
)
(0)
TODO: how long would this be?
So we see that all of these rules resolve quite quickly and do not go into each other. S however offers some problems, in that:
C_0 = Z
C_i = S(C_{i-1})
D_i = C_i(S)(S)
So we see that D_i goes somewhat simply into C_i, and C_i is recursive giving:
S^i(Z)
Calculate the nine first digits of:
D_a(D_b)(D_c)(C_d)(A)(e)
Removing D_a:
S^i(Z)S)(S)(D_b)(D_c)(C_d)(A)(e)
A naive T in Python is:
from collections import deque

def T(a: int, b: int, N: int) -> int:
    total = a
    q = deque([a] * (a - 1))
    is_a = False
    for i in range(N - 1):
        cur = q.popleft()
        total += cur
        q.extend([a if is_a else b] * cur)
        is_a = not is_a
    return total

assert T(2, 3, 10) == 25
assert T(4, 2, 10**4) == 30004
assert T(5, 8, 10**6) == 6499871
which passes the tests, but takes half a second on PyPy. So clearly it is not going to work for 22332223332233 which has 14 digits.
Maybe if T is optimized enough, then we can just bruteforce over the ~40k possible sum ranges 2 to 223. 1 second would mean 14 hours to do them all, so bruteforce but doable. Otherwise another optimization step may be needed at that level as well: one wonders if multiple sums can be factored out, or if the modularity can of the answer can help in a meaningful way. The first solver was ecnerwala using C/C++ in 1 hour, so there must be another insight missing, unless they have access to a supercomputer.
The first idea that comes to mind to try and optimize T is that this is a dynamic programming, but then the question is what is the recurrence relation.
The sequence appears to be a generalization of the Kolakoski sequence but to numbers other than 1 and 2.
maths-people.anu.edu.au/~brent/pd/Kolakoski-ACCMCC.pdf "A fast algorithm for the Kolakoski sequence" might provide the solution, the paper says:
It is conjectured that the algorithm runs in time and space , where
and provides exactly a recurrence relation and a dynamic programming approach.
www.reddit.com/r/algorithms/comments/8cv3se/kolakoski_sequence/ asks for an implementation but no one gave anything. Dupe question: math.stackexchange.com/questions/2740997/kolakoski-sequence contains an answer with Python and Rust code but just for the original 1,2 case.
github.com/runbobby/Kolakoski has some C++ code but it is not well documented so it's not immediately easy to understand what it actually does. It does appear to consider the m n case however.
Numerical solution:
166666666689036288
Programs:
GPT-5 produced some C++ code, we told it it was wrong and second try worked:
29337152.09
Programs:
Numerical solution:
357591131712034236
As mentioned at euler.stephan-brumme.com these tend to be harder, as they have their own judge system that actually runs programs, and therefore can test input multiple test cases against their reference implementation rather than just hard testing the result for a single input.
Goes only up to Project Euler problem 254 as of 2025, which had been published much much earlier, in 2009, so presumably they've stopped there.

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