Enigmatic Code

Programming Enigma Puzzles

Enigma 1378: Exaudi Deus

From New Scientist #2538, 11th February 2006

In our church we use a hymnbook with fewer than a thousand hymns. On Sunday we sing an opening hymn, a psalm, and three more hymns, the numbers being displayed in that order on the board.

Recently, as I sat at the organ I noticed that the five numbers were all different perfect squares in increasing order, and that each digit which did occur, occurred exactly three times.

What was the number of the psalm?



One response to “Enigma 1378: Exaudi Deus

  1. Jim Randell 29 October 2013 at 8:55 am

    With such a relatively small number of candidate squares to consider it’s easy just consider all possible combinations, which gives a compact Python program that runs in 339ms.

    from itertools import combinations
    from collections import Counter
    from enigma import irange, printf
    # squares less than 1000 (as strings)
    squares = list(str(i * i) for i in irange(1, 31))
    # choose 5 squares
    for s in combinations(squares, 5):
      t = ''.join(s)
      # the number of digits must be a multiple of 3
      if len(t) % 3: continue
      # count the digits
      c = Counter(t)
      # each one must appear 3 times
      if any(v != 3 for v in c.values()): continue
      printf("psalm = {p} [{s}]", p=s[1], s=', '.join(s))

    Solution: The psalm was number 64.

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