S2T2

Jim Randell 10:05 am on 27 January 2026 Permalink Reply
Tags: by: Derek Emley   

• 

  Jim Randell 10:06 am on 27 January 2026 Permalink | Reply

  If we suppose the original number n is of the form:

  n = 10a + z
  where z is the final digit, and a is a k-digit number

  Then moving the final digit to the front gives us:

  2n = (10^k)z + a

  From which we derive that a is a k-digit number, such that:

  19a = (10^k − 2)z

  If the number contains no digit more than twice, then the longest the number can be is 20 digits long, and so we can consider k up to 19.

  This Python program runs in 70ms. (Internal runtime is 719µs).

  from enigma import (irange, div, nsplit, rotate, zip_eq, seq2str, printf)
  
  # check a candidate number <n> (with digits <ds>)
  def check(n, ds):
    # check no digit occurs more than twice
    if any(ds.count(x) > 2 for x in set(ds)): return
    # record shifts that result in doublings
    ns = [n]
    # collect doubling sequences
    while True:
      n *= 2
      ds = rotate(ds, -1)
      if not zip_eq(ds, nsplit(n)): break
      ns.append(n)
    return ns
  
  # look for up to 19 digits prefixes
  for k in irange(1, 19):
    x = 10**k - 2
    # consider possible final digits
    for z in irange(1, 9):
      a = div(x * z, 19)
      if a is None: continue
      n = 10*a + z
      ds = nsplit(n)
      if len(ds) != k + 1: continue
      ns = check(n, ds)
      if ns and len(ns) >= 4:
        printf("{ns}", ns=seq2str(ns, sep=" -> ", enc=""))
  

  Solution: The number is: 105263157894736842.

  The number is 18 digits long, and uses each digit twice except 0 and 9 (which are used once each).

  Shifting the final digit to the front multiple times we get:

  105263157894736842 = n
  210526315789473684 = n × 2
  421052631578947368 = n × 4
  842105263157894736 = n × 8

  A run of 4 is the longest run we can get in base 10 (as n × 16 will not have the same number of digits as n), but there is another run of 3 we can make:

  157894736842105263 = n
  315789473684210526 = n × 2
  631578947368421052 = n × 4

  These numbers are the repetends of fractions of the form k / 19.

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• 

  Frits 8:51 am on 31 January 2026 Permalink | Reply

  '''
    n = xbcd where x is a k-digit number and b,c and d are digits 
  2.n = dxbc
  4.n = cdxb
  8.n = bcdx with x, b and c all even
  
  8.n = bcdx also has k+3 digits so b = 8 and first digit of x = 1
     --> c = 4 (not 9) and last digit of x must be 6
         d = 2 (as 2.n must be 2.....)
         
  n = 1....6842      
  
    n = xbcd = 1000.x + bcd  (x has k digits)
  8.n = bcdx = 10^k . bcd + x
  
  8000.x + 8.bcd = 10^k . bcd + x
  7999.x = 19.421.x = (10^k - 8).842
  19.x = 2.(10^k - 8)
  '''
        
  for k in range(2, 18):  
    x, r = divmod(2 * (10**k - 8), 19)
    if r: continue
    n = str(x) + "842"
    # no digit occurs more than twice
    if all(n.count(str(i)) <= 2 for i in range(10)): 
      print("answer method 1:", n)
    
  # in order for 2 * (10**k - 8) % 19 = 0 we must have
  # 2 * 10**k % 19 = 16 
  # 2 * 10**k = 19.10**(k - 1) + 10**(k - 1)
  # so 10**(k - 1) % 19 = 16
  
  # 10**(k - 1) % 19 follows a logical pattern
  k = 2
  a = 10**(k - 1) % 19
  while a != 16: 
    a = (a + 19 * (a % 2)) // 2
    k += 1
  
  n = str(2 * (10**k - 8) // 19) + "842"
  # no digit occurs more than twice
  if all(n.count(str(i)) <= 2 for i in range(10)): 
    print("answer method 2:", n)
  

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