S2T2

Jim Randell 9:16 am on 20 January 2026 Permalink Reply
Tags: by: Nick MacKinnon ( 52 )   

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  Jim Randell 9:17 am on 20 January 2026 Permalink | Reply

  Programatically, it is straightforward to just try all possible bases.

  This Python program runs in 75ms. (Internal runtime is 4.2ms).

  from enigma import (irange, mlcm, call, nconcat, printf)
  
  # M = LCM(2..16)
  M = call(mlcm, irange(2, 16))
  
  # consider possible bases
  for b in irange(2, 9999):
    # n = 1111 [base b]
    n = nconcat([1, 1, 1, 1], base=b)
    # check for divisibility by 2..16
    if n % M == 0:
      # output solution
      printf("base={b} n={n}")
  

  But here is a more sophisticated approach:

  Another way to write 1111 [base b] is: b^3 + b^2 + b + 1, which is an integer valued polynomial function.

  And this means we can use the [[ poly_roots_mod() ]] solver from the enigma.py library (described here: [ link ]) to solve the puzzle.

  The following Python program has an internal runtime of 572µs.

  from enigma import (Polynomial, rev, irange, mlcm, call, poly_roots_mod, printf)
  
  # calculate 1111 [base b]
  p = Polynomial(rev([1, 1, 1, 1]))
  
  # M = LCM(2..16)
  M = call(mlcm, irange(2, 16))
  
  # find roots of p(b) = 0 [mod M] that give a valid base
  roots = poly_roots_mod(p, 0)
  for b in roots(M):
    if 1 < b < 10000:
      # calculate the corresponding n value
      n = p(b)
      # output solution
      printf("base={b} n={n}")
  

  Solution: The base is: 5543.

  And the round number is: 170338568400 = 720720 × 236345.

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  Ruud 10:52 am on 20 January 2026 Permalink | Reply

  import istr
  
  print(*(base for base in istr.range(2, 10000) if all(sum(base**i for i in range(4)).is_divisible_by(i) for i in range(2, 17))))
  

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