|Re: How to speak the problem?|
Message #19 Posted by Dave Shaffer (Arizona) on 21 June 2007, 11:09 p.m.,
in response to message #18 by Allen
"4. i must be 3 (only CD with 126 and 12 (step 2))"
No - a 6 works, too (to give Olivier's solution)!
(I had already prepared the following):
Actually, if we read between the lines, there ARE other constraints.
If this is a sudoku of the normal type, perhaps the most compelling constraint is that the values must be (positive) integers. And, as noted by others, they must lie between 1 and 9 and in fact include ALL the integers from 1 to 9.
So, letís try some educated guess work. First, note that abc=96, def=315, and cfi=126 tell us that none of a, b, c, d, e, f, i can be a 1. (If one of the variables in a triple product was one, the
maximum product would be 8x9=72.) Thus, we know already that either g or h must be 1. ghi=12 tells us that the other two (the pair g and i or the pair h and i) must be either the pair 2 and 6 or the pair 3 and 4. (Turns out, both work.)
Now, since beh = 40 and e=5, we know that bh = 8. If b and h are positive integers (between 1 and 9!), then b and h must be the pair 1 and 8 or the pair 2 and 4. Similarly, def = 315 leads to the conclusion that df = 63 and thus d and f are the pair 7 and 9. Since adg = 72, d must be a 9 (and f a 7), because if d was a 7, it would not divide evenly into 72.
We now have for sure: e = 5 (given), d = 9, and f = 7.
Also, from adg = 72, and d = 9, we get ag = 8. We already ascertained that g must be 1, or 2 or 6, or 3 or 4. Since all numbers must be integers, ag = 8 rules out g = 6 or 3, so g must be 1 or 2 or 4.
Note, too, that cfi = 126 gives ci = 18 (because f = 7). So, c and i are either the pair 2 and 9 or the pair 3 and 6. Because d = 9, c and i must be the pair 3 and 6.
At this point, try b = 1 or 2 or 4 or 8, using abc = 96. Look for inconsistencies with the known values. Do the same for other possible pairs.
When you are done, I think you will find that there are at least two solutions consistent with all the triple product conditions:
a b c d e f g h i = 8 4 3 9 5 7 1 2 6 (as reported by Olivier)
a b c d e f g h i = 8 2 6 9 5 7 1 4 3 (my alternate solution)
Edited: 21 June 2007, 11:09 p.m.