If the number is abc, then you are told
a+b+c = 20
b = (a+c)/2
100c+10b+a = 100a+10b+c + woo
Now just solve for a,b,c
a+b+c = 20
b = (a+c)/2
100c+10b+a = 100a+10b+c + woo
Now just solve for a,b,c
Let's assume the original three-digit number is represented as ABC, where A represents the hundreds digit, B represents the tens digit, and C represents the ones digit.
1. The sum of the digits of the three-digit number is 20.
This can be represented as: A + B + C = 20.
2. The middle digit is equal to one-fourth of the sum of the other two digits.
This can be represented as: B = (A + C) / 4.
3. Reversing the order of the digits increases the original number by 100A + 10B + C.
The reversed number will be CAB, and the increase can be represented as: 100A + 10B + C - (100C + 10A + B) = 100A + 10B + C - 100C - 10A - B = 99A + 9B - 99C = 99(A - C) + 9(B + C) = 200.
Now, we will solve these equations to find the values of A, B, and C, which will give us the original number.
From the first equation A + B + C = 20, we can rewrite it as A + C = 20 - B and substitute it in the second equation:
B = (A + C) / 4
B = (A + (20 - B)) / 4
4B = A + 20 - B
5B = A + 20
Similarly, from the third equation, we can rewrite it as:
99(A - C) + 9(B + C) = 200
11(A - C) + B + C = 22
11(A - 20 + B) + B + 20 = 22
11A - 220 + 11B + B = 22
11A + 12B = 242
We now have two equations and two variables; we can solve this system of equations using substitution or elimination method. Let's use the substitution method:
From the equation 5B = A + 20, we can rewrite it as A = 5B - 20.
Substituting this value in the equation 11A + 12B = 242:
11(5B - 20) + 12B = 242
55B - 220 + 12B = 242
67B = 462
B = 6
Substituting this value of B back into A = 5B - 20:
A = 5(6) - 20
A = 10
Now, we can substitute the values of A and B into the equation A + B + C = 20:
10 + 6 + C = 20
C = 4
Therefore, the original three-digit number is 106.
Hope this helps!