Arrange the digits 1-9 in a 3x3 square using each once only according to the following rules:
- Exactly one prime is directly above a prime two less.
- Each pair of opposite corners sums to the same square total and exactly two columns share another square total.
- Exactly one prime is directly to the left of a non-prime two greater.
- Directly above exactly one square digit is a digit four greater.
- Directly to the right of exactly one cubic digit is a digit one greater.
This one requires neither a program nor time-consuming deductive thought. If you focus at first on conditions 1,3,4 and 5, there are basically 8 conditions in four mutually exclusive pairs. Call them s,t,u,v,w,x,y,z.
(s) 5/3 or (t) 7/5;
(u) 24 or (v) 79;
(w) 8/4 or (x) 5/1;
(y) 12 or (z) 89;
There are 2^4=16 groups of possible combinations. So you just evaluate each possible matrix until you get one that meets rule 2. This is actually possible to do fairly quickly, so brain-straining ratiocination is uncalled for.
(s u w y) 5/3 24 8/4 12 It is impossible to satisfy this combination in a 3x3 matrix.
(s u w z) 5/3 24 8/4 89, with 1,6,7 unaccounted for.
The only possible pattern is:
?89
245
??3
Fails rule 2.
etc. etc..... until we get to:
(t u x z) 7/5 24 5/1 89 with 3,6 unaccountted for. There are 12 possible patterns here:
724 789 7?? 7?? 724 789
589 524 524 589 5?? 5??
1?? 1?? 189 124 189 124
247 897 ??7 ??7 247 897
895 245 245 895 ??5 ??5
?11 ??1 891 241 891 241
Only the last one can possibly satisfy rule 2. We are down to two possibilities:
897 897
365 635
241 241
Only the second one meets both stipulations of rule 2.
The answer is:
897
635
241
Edited on December 29, 2003, 2:01 pm
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Posted by Penny
on 2003-12-29 13:55:49 |
Solution (no program needed)
