Originally posted by ^tamago^:
1ab
cd7
efg
1+7+a+b+c+d+e+f+g = 45
a+b+c+d+e+f+g = 37
1+a+c+d = a+b+d+7 = c+d+e+f = d+f+g+7
1+a+c = a+b+7 = c+e+f = f+g+7
==> 4 equations, 6 vars, unable to solve in real no. set,
listing out relations:
c+1 = b+7
c+e = g+7
a+1 = e+f
a+b = f+g
==> found: b = c-6
for b,c is member of Z+ such that 1<=x<=9,
b is either 1,2,3
c is either 7,8,9
==> 7 and 1 taken
==> scenario split
scenario 1) b=2 and c=8
or
scenario 2) b=3 and c=9
SCENARIO 1 (b=2 and c=8 )
==> 1a2
==> 8d7
==> efg
a+d+9 = d+e+f+8 = d+f+g+7
==> g = e+1
==> a = e+f-1
==> 1 e+f-1 2
==> 8 d 7
==> e f e+1
==> undefined numbers 3,4,5,6,9
==> need to define d,e,f through trial and error
==> e+f-1 <= 9
==> e+f <= 10
==> e != 6 (g cannot be 7)
==> f != e+1
==> truncate e=3,f=4; e=4,f=5; e=5,f=6
==> e+f-1 != 7
==> truncate e=3,f=5; e=5,f=3
==> e+f-1 != 8
==> truncate e=3,f=6; e=5,f=4
==> listing solution(s)
e f a g d
4 6 9 5 3
[b]192
837
465
==> 1 solution found under SCENARIO 1
SCENARIO 2 (b=3 and c=9)
==> 1a3
==> 9d7
==> efg
a+d+10 = d+e+f+9 = d+f+g+7
==> g = e+2
==> a = e+f-1
==> 1 e+f-1 3
==> 9 d 7
==> e f e+2
==> undefined numbers 2,4,5,6,8
==> need to define d,e,f through trial and error
==> e+f-1 <= 9
==> e+f <= 10
==> e != 5 (g cannot be 7)
==> g != f+2
==> truncate e=2,f=4; e=4,f=6; e=6,f=8
==> e+f-1 != 7
==> truncate e=2,f=6, e=6,f=2
==> e+f-1 != 9
==> truncate e=2,f=8; e=8,f=2; e=4,f=6; e=6,f=4
==> listing solution(s)
e f a g d
2 5 6 4 8
4 5 8 6 2
163
987
254
183
927
456
==> 2 solutions found under SCENARIO 2
dunno if this helps. just playing around with algebra.
[/b]
thanks very much! although i do not really understand your
scenario 2...you use maths calculator izzit? some parts i really
dunno how you get... but really thanks! you are darn smart! ^^
