two solid position

TWO SOLID POSITION The size of numbers are from one through 39 we have 10 numbers 0 1 2 3 4 5 6 7 8 9 Input A: ( delete 2 or more numbers) example :4 & 6 (04,14,24,34 & 06 16,26,36) = 8 numbers total ,out of basket. so the active numbers are: 1 2 3 5 7 8 9 10 11 12 13 (15) (17 18 19 20 21 22 23 25) 27 28 29 30 31 32 33 35 37 38 39 Input B: choose of 2 numbers , for example 05 25 we have 10 options, we need to develop: AB=15 25, AC= 15 25, AD= 15 25 AE= 15 25, BC=15 25, BD=15 25, BE= 15 25 CD=15 25 CE=15 25 , DE=15 25 Input C: choose one : BC , it means 15 25 will be solid on that position X 15 25 X X ( so any number in between will be eliminated, automatically) (17,18,19,20,21,22,23,) total of 7 will be eliminated automatically. Because 15 +25= 40 is occupied in position B and C Input D: type Sum: 119(it is given) It means if the answer is 05+15+25+35+39 = 119 So the remainder is (119-40) = 79 will be the result of ( A+DE) Therefore : position (A) is ( 1 2 3 5 6 7 8 9 10 11 12 13), one will be chosen Therefore, position DE ( 27 28 29 30 31 32 33 35 37 38 39), 2 will be chosen Output: (15+25) is solid in BC + one number for position A and two numbers for position DE The answer is ( 05 15+ 25 +35 + 39)= 119 so (5+35+39=79) Program code will calculate One of A and two of DE to be equal to 79 then add 40=119 Output: Print in notepad and give a name C# language