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Sunday Times Teaser 2820 – Three Ages

by Andrew Skidmore

Published: 9 October 2016 (link)

Today is Alan’s, Brian’s and Colin’s birthday. If I write down their ages in a row in that order then I get a six-figure number. If I write down their ages in a row in the reverse order (ie, Colin’s followed by Brian’s followed by Alan’s) then I get a lower six-figure number. When I divide the difference between these two six-figure numbers by the total of their three ages the answer is Alan’s age multiplied by Colin’s.

What are Alan’s, Brian’s and Colin’s ages?

2 Comments Leave one →
  1. Brian Gladman permalink

    Here is a simple solution that depends on an assumption that the three ages each have two digits (this seems to be necessary for a reasonable manual solution).

    As other have found here, this teaser has a flaw since it has two reasonable solutions.

    There are five possible solutions when we don’t restrict the ages in any way: (A, B, C) = (22, 61, 18), (99, 70, 33), (78, 252, 6), (37, 326, 3) and (51, 830, 3).

  2. For the case where all three ages are two digits (AB, CD, and EF) we can use the general alphametic solver from the library to get a one-line solution.

    This runs in 206ms (using PyPy).

    Other cases can be investigated similarly.

    The library is available at [ ].

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