Posted by Mohammed on Sunday, March 21, 2010 at 3:38am.
One digit numbers: 6 (one possibility only)
Two digit non-zero numbers:
15, 24, 33, 42 and 51 (total = 5)
Three digit numbers:
123, 213, 132, 312, 321, 231
114, 411, 141, 222 (total = 10)
Four digit numbers:
1113, 1131, 1311, 3111
1221, 2121, 1122, 2211
2112, 1212 (total = 10)
Five digit numbers:
11112, 11121, 11211, 12111, 21111
(5 numbers)
Six digit numbers: 111111 (one only)
Total number of possibile arrangements = 32 (without using zero)
Sum of 1: One possibility (1)
Sum of 2: Two possibilites (2,11)
Sum of 3: Four possibilities
(111, 12, 21, 3)
Sum of 4: (Eight possibilities.
(4, 13, 31, 22, 112, 121, 211, 1111)
Thus for a sum of five there are 16 possibilities.
For a sum of N there are 2^(N-1) possibilities.
How to solve x+7y=8 8x-4y=-2