Re: Project Euclid Problem 26 SOLVED!!!

On 3/9/2024 7:28 AM, Farley Flud wrote:

Project Euclid Problem is SOLVED!!!
 https://projecteuler.net/problem=26
 Using the fantastically potent FOSS CAS Maxima/wxMaxima the solution
is a fucking snap:
 max:1$
maxd:3$
for num:3 thru 1000 step 1 do (
     d:1,
     fact_list:ifactors(num),
     for i:1 thru length(fact_list) do (
     if not(fact_list[1] = 2 or fact_list[1] = 5) then d:d*fact_list[1]^fact_list[2]
       ),
     if not(d = 1) then (
         rep:zn_order(10,d), print("denom = ", num, "repeats = ", rep),
         if (rep > max) then (max:rep, maxd:num)
     )
     else (print("denom = ", num, "terminating"))
);
print("Max rep =", max, "at denom =", maxd);
  The output is not ideal, but hey, I am a fucking messy eater.
I spill my coffee.  I slobber all over my shirt.  But I ALWAYS
get the correct answer in the most efficient way and that's
ALL that matters.
 Hail Linux!
 Hail GNU, FOSS, and the FSF!
 Hail Stallman!
 

I don't understand. You say it's solved. So what did your code find? Which denominator under 1000 gives the longest repeating decimals if numerator is 1?
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