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Sunday Times Teaser 2884 – Farewell

by Victor Bryant

Published December 31 2017 (link)

Today I am retiring after editing this column for 40 very pleasurable years. My heartfelt thanks go out to all the setters and solvers of puzzles with whom I have corresponded.

To celebrate the 40 years I threw a party for the puzzle-setters. At the party we assigned a different whole number from 1 to 26 to each letter of the alphabet; for example, we had A=13 and Z=3. We did this in such a way that, for each person present (including me), the values of the letters of their Christian name added to 40. Bob, Graham, Hugh, Nick and Richard were there, as were two of Andrew, Angela, Danny, Des, Ian, John, Mike, Peter, Robin, Steve and Tom.

Which two?

2 Comments Leave one →
  1. Brian Gladman permalink

    There are twelve unique assignments of values to letters: A = 13, B = 16, D = 10, E = 18, G = 7, H = 4, M = 2, O = 8, R = 1, S = 12, U = 25 and Z = 3. There are then three interchangeable assignment pairs (C, I) = (5, 6), (K, N) = (14, 15) and (T, V) = (9, 11). Finally J can be any of eight values: 17, 19, 20, 21, 22, 23, 24 or 26. This would give 64 solutions but if (C, I) are interchanged then so are (K, N), which reduces the number of solutions to 32.

  2. I treated this puzzle as a pair of linked alphametic problems in base 27, using the SubstitutedExpression() solver from the library.

    It’s not especially quick, but quite succinct.

    The code can be executed on [ ].

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