On 12/8/18 12:40 PM, Avi Gross wrote: > You are solving for: ab + aa + cd == ce
Actually, an even quicker analysis for this particular problem is: from the 10s digits, a + a + c + carryin = c Thus a and carryin must both be 0 (carryin can not be negative, nor any of the variables) thus the final solution space is: b + d = e a = 0 c any other digit (6 possible values for every combo of b, d, e) if b is <= 4, there are 8-b possible values of d that will have a legal value of e. b = 1, we get d = 2, 3, 4, ... 7, 8 b = 2, we get d = 1, 3, 4, ... 6, 7 (8 would generate carry) b = 3, we get d = 1, 2, 4, 5, 6 b = 4, we get d = 1, 2, 3, 5 if b >= 5 we get 9-b possible values of d (we no longer have to omit the possible value of b = d) So the number of possible answers are: (7+6+5+4+4+3+2+1)*6 = 192 (your 320 was you gave c 10 possible values, but you need to remove the duplicates). -- https://mail.python.org/mailman/listinfo/python-list
