New Scientist Enigma 936 – Shift of Fortune
by Susan Denham
From Issue #2091, 19th July 1997 (link)
To enter the lottery I chose six lucky numbers: I called this “selection A”. I have added 1 to each of those six numbers to form a new “selection B”, I’ve added 1 again to each to form “selection C”, and repeated this twice more to form “selection D” and “selection E”. So, for example, the lowest number in selection E is four more than the lowest number in selection A. Of course, as always, each selection is in the range 1-49.
In one of the selections (but not selection A) the six numbers are all two-digit numbers with the same first digit.
In one of the selections (but not selection A) there are no prime numbers.
Each week I decide to have two goes at the lottery and I choose at random two of my five selections. I know that whichever pair I choose, it would be possible for one of the selections to contain four of the six winning numbers and the other selection to contain none. On the other hand, whichever pair I choose, it would also be possible for both selections to contain four of the six winning numbers.
Which six numbers form “selection A”?