Enigmatic Code

Programming Enigma Puzzles

Enigma 1188: Square roundabout

From New Scientist #2344, 25th May 2002 [link]

I was playing about with my seven-digit-display calculator, showing my numerate nephew a trick or two. I displayed a number on the calculator and he looked at it upside down. After some jottings of his own he declared “I can see a number too, and it’s a perfect square”.

I then doubled my original number and displayed the answer and again he looked at it upside down and did some calculations.

“I can still see a number, and it’s another perfect square”, he said.

Which number did I originally display?



One response to “Enigma 1188: Square roundabout

  1. Jim Randell 21 December 2015 at 8:34 am

    This Python program runs in 53ms.

    from itertools import count
    from enigma import is_square, join, irange, printf
    # numbers which read as numbers when inverted on a 7-segment display
    inverted = { '0': '0', '1': '1', '2': '2', '5': '5', '6': '9', '8': '8', '9': '6' }
    # invert a number (passed as a decimal string)
    def invert(n):
        s = join(inverted[x] for x in reversed(n))
      except KeyError:
        return None
      i = int(s)
      return (i if str(i) == s else None)
    # consider increasing squares for the first inverted number
    for r1 in count(1):
      i1 = str(r1 * r1)
      # it can't have more than 7 digits
      if len(i1) > 7: break
      # invert it to get the original number
      n1 = invert(i1)
      if n1 is None: continue
      # double the original number
      n2 = str(n1 * 2)
      if len(n2) > 7: continue
      # and invert it
      i2 = invert(n2)
      if i2 is None: continue
      # that should also be a square
      r2 = is_square(i2)
      if r2 is None: continue
      printf("first number = {n1}, inverted = {i1} ({r1}^2), second number = {n2}, inverted = {i2} ({r2}^2)")

    Solution: The original number was 9126 (the correct way up).

    9126 reads as 9216 (= 96²) when inverted.

    9126 doubled is 18252, which reads as 25281 (= 159²) when inverted.

    We ignore the trivial solution where all numbers are 0, because there is some calculation involved to determine if a number is a perfect square.

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