Python

# snipped read_input implementation
data = read_input("input_p3.txt")

# part 1 and 2 can be solved using a calculator 
# with only the first and last gears

def part3(data):
    shafts = data.splitlines()

    rotations = 100 * int(shafts[0])
    for shaft in shafts[1:-1]:
        gear1, gear2 = [int(g) for g in shaft.split('|')]
        rotations *= gear2 / gear1
    rotations /= int(shafts[-1])
    
    return int(rotations)

assert part3("""5
7|21
18|36
27|27
10|50
10|50
11""") == 6818

print(part3(data))

Uiua

Fortunately the ratios for the common gears all proved to be rational in part 3.

⌊×2025/÷⊏¯1_0[128 64 32 16 8]              # ->32400
⌈×⊙÷°⊟⊏¯1_0⊙10000000000000[128 64 32 16 8] # -> 625000000000
⊜□⊸≠@\s "5 5|10 10|20 5"
⌊×100×⊃(÷∩⋕⊃⊣⊢|/×≡◇(/÷⊜⋕⊸≠@|)↘1↘¯1) # -> 400

I liked this one!

import Control.Arrow  
import Control.Monad  
import Data.List  
import Data.Ratio  

simpleTrain = uncurry (%) . (head &&& last) . map read  

compoundTrain input =  
  let a = read $ head input  
      z = read $ last input  
      gs =  
        map  
          ( uncurry (%)  
              . (read *** read . tail)  
              . break (== '|')  
          )  
          $ (tail . init) input  
   in foldl' (/) (a % z) gs  

part1, part2, part3 :: [String] -> Integer  
part1 = floor . (2025 *) . simpleTrain  
part2 = ceiling . (10000000000000 /) . simpleTrain  
part3 = floor . (100 *) . compoundTrain  

main =  
  forM_  
    [ ("everybody_codes_e2025_q04_p1.txt", part1),  
      ("everybody_codes_e2025_q04_p2.txt", part2),  
      ("everybody_codes_e2025_q04_p3.txt", part3)  
    ]  
    $ \(input, solve) -> readFile input >>= print . solve . lines  

maaath

Scheme/Guile

(import (rnrs io ports (6)))

(define (parse-file file-name)
       (map string->number (string-split (call-with-input-file file-name get-string-all) #\newline)))

(let* ((gears (parse-file "notes/everybody_codes_e2025_q04_p1.txt")))
  (format #t "P1 Answer: ~a\n\n" (* 2025 (/ (car gears) (car (last-pair gears))))))


(let* ((gears (parse-file "notes/everybody_codes_e2025_q04_p2.txt")))
  (format #t "P2 Answer: ~a\n\n" (ceiling (* 10000000000000 (/ (car (last-pair gears)) (car gears))))))


(define (parse-file-p3 file-name)
       (map
              (lambda (line) (map string->number(string-split line #\|)))
              (string-split (call-with-input-file file-name get-string-all) #\newline)))
(let* ((gears (parse-file-p3 "notes/everybody_codes_e2025_q04_p3.txt")))
  (format #t "P2 Answer: ~a\n\n"
          (floor (* 100
             (apply * (map (lambda (gear) (if (= 1 (length gear)) 1 (/ (cadr gear) (car gear)))) gears))
             (/ (caar gears) (caar (last-pair gears)))))))

In 1980 we didn’t have tuple or set or vector types. We addressed elements of our cons-based data structures with nonsense incantations like caddar and we liked it that way.

(ql:quickload :cl-ppcre)

(defun read-inputs (filename)
  (mapcar #'parse-integer (uiop:read-file-lines filename)))

(defun main-1 (filename)
  (let ((gears (read-inputs filename)))
    (truncate (* 2025 (/ (first gears) (car (last gears)))))))

(defun main-2 (filename)
  (let ((gears (read-inputs filename)))
    (ceiling (* 10000000000000 (/ (car (last gears)) (first gears))))))

(defun read-inputs-3 (filename)
  (flet ((parse-line (line)
           (or (ppcre:register-groups-bind
                   (a b)
                   ("^([0-9]+)[|]([0-9]+)$" line)
                 (cons (parse-integer a) (parse-integer b)))
               (let ((a (parse-integer line)))
                 (cons a a)))))
    (mapcar #'parse-line (uiop:read-file-lines filename))))

(defun gear-ratio-3 (gears)
  (labels ((iter (gears ratio)
             (if (< (length gears) 2)
                 ratio
                 (let ((out-gear (cdar gears))
                       (in-gear (caadr gears)))
                   (iter (cdr gears) (* ratio (/ out-gear in-gear)))))))
    (iter gears 1)))

(defun main-3 (filename)
  (let ((gears (read-inputs-3 filename)))
    (truncate (* 100 (gear-ratio-3 gears)))))

Rust

use num::{BigInt, Integer};

pub fn solve_part_1(input: &str) -> String {
    let gears: Vec<i64> = input.trim().lines().map(|g| g.parse().unwrap()).collect();
    (2025 * gears[0] / gears.last().unwrap()).to_string()
}

pub fn solve_part_2(input: &str) -> String {
    let gears: Vec<i64> = input.trim().lines().map(|g| g.parse().unwrap()).collect();
    let res = (BigInt::parse_bytes(b"10000000000000", 10).unwrap() * gears.last().unwrap())
        .div_ceil(&(BigInt::ZERO + gears[0]));
    res.to_string()
}

pub fn solve_part_3(input: &str) -> String {
    let mut lines = input.trim().lines();
    let first_gear = BigInt::parse_bytes(lines.next().unwrap().as_bytes(), 10).unwrap();
    let mut nominator: BigInt = first_gear * 100;
    let mut denominator: BigInt = BigInt::ZERO + 1;
    for line in lines {
        let mut split = line.split("|");
        denominator *= BigInt::parse_bytes(split.next().unwrap().as_bytes(), 10).unwrap();
        match split.next() {
            Some(size) => {
                nominator *= BigInt::parse_bytes(size.as_bytes(), 10).unwrap();
            }
            None => {
                break;
            }
        }
    }
    (nominator / denominator).to_string()
}

Nim

For part 3 I parse gears as tuples, with regular gears having same value on both ends e.g.

3|5 -> (3, 5)
3   -> (3, 3)

proc parseGears(input: string): seq[int] =
  for line in input.splitLines():
    result.add parseInt(line)

proc parseNestedGears(input: string): seq[(int, int)] =
  for line in input.splitLines():
    let nested = line.split('|').mapIt(it.parseInt)
    result.add:
      if nested.len == 1: (nested[0], nested[0])
      else: (nested[0], nested[1])

proc solve_part1*(input: string): Solution =
  let gears = parseGears(input)
  result := 2025 * gears[0] div gears[^1]

proc solve_part2*(input: string): Solution =
  let gears = parseGears(input)
  result := ceil(10000000000000'f64 / (gears[0] / gears[^1])).int

proc solve_part3*(input: string): Solution =
  let gears = parseNestedGears(input)
  let ratios = (0..gears.high-1).mapIt(gears[it][1] / gears[it+1][0])
  result := int(100 * ratios.prod)

Full solution at Codeberg: solution.nim