# 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 →

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).

For the case where all three ages are two digits (AB, CD, and EF) we can use the general alphametic solver from the

enigma.pylibrary to get a one-line solution.This runs in 206ms (using PyPy).

Other cases can be investigated similarly.

The

enigma.pylibrary is available at [ http://www.magwag.plus.com/jim/enigma.htlm ].