Enigmatic Code

Programming Enigma Puzzles

Enigma 1523: Dicey numbers

From New Scientist #2685, 6th December 2008 [link]

I threw a small number of standard dice and noted that the grand total of spots on the tops was a perfect square. Then I put the dice in a row in increasing order of their tops and wrote down the large number so formed (for example, 12224…).

Then I repeated the exercise with double the number of dice. Again the total number of spots was a perfect square, and again I wrote down the number formed by the tops in increasing order. It turned out that the second number I wrote down was the square of the first.

Incidentally (although you do not need to know this) throughout the exercise I threw at most one 6.

What was the second number that I wrote down?


One response to “Enigma 1523: Dicey numbers

  1. Jim Randell 30 July 2012 at 7:51 am

    The following Python code runs in 100ms.

    It checks for solutions up to a specified “small number” of dice. The default is 10, but you can specify a different number on the command line. If you want to find only the first solution you can add a call to [[ sys.exit() ]] at the end of the [[ check() ]] function.

    import re
    import sys
    from enigma import irange, is_square, printf
    # check to see if this sequence is a solution
    # s - digit sequence as a string
    def check(s):
      # the sum of the digits must be a square
      t = sum(int(i) for i in s)
      if not is_square(t): return
      # square the number
      s2 = str(int(s) ** 2)
      if len(s2) != 2 * len(s): return
      # the number must be formed from increasing dice (with at most one 6)
      if not re.match(r'^1*2*3*4*5*$' if '6' in s else r'^1*2*3*4*5*6?$', s2): return
      # again, the sum of the digits should be a square
      t2 = sum(int(i) for i in s2)
      if not is_square(t2): return
      printf("[{n}] {s} => {t}, {s2} => {t2}", n=len(s))
      # sys.exit(0) # if you only want the first answer
    # construct n-digit sequences of increasing dice digits (with just one 6)
    # n - number of digts we're looking for
    # s - current string of digits
    # d - final digit (as an integer)
    def solve(n, s='', d=1):
      if len(s) == n:
      if d == 6:
      for i in irange(d, 6):
        solve(n, s + str(i), i)
    MAX = 10 if len(sys.argv) < 2 else eval(sys.argv[1])
    for n in irange(1, MAX):

    Solution: The second number is 1111111155555556.

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