==> pickover/pickover.07.p <==
Title: Cliff Puzzle 7: 3x3 Recursion
From: cliff@watson.ibm.com

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information.  PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup.  I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise.  Thanks, Cliff Pickover

* * *

Consider the 3x3 array below.  All nine digits are used exactly once.

1 9 2
3 8 4
5 7 6

Notice that "384" is twice the number in the first row, and that
"576" is three times the number in the first row.

Questions:
1.  Are there other ways of arranging the number to produce the same
result using each digit only once and the same rules?
Remember, the second row must be twice the first.  The third row
must be 3 times the first row.

2.  Start with the number in the last row (e.g "576" or any other
solution you may find) and continue to form another 3x3 matrix using the
same rules with the new starting number.  In other words, the number in
the second row must be twice the first.  The third row must be three
times the first.  (For this problem you may truncate any digits in the
beginning.  For example, 1384 would become 384.)

Keep going.  How many matrices can you create before it is impossible
to continue.  Again, each digit must be used only once
in each matrix.