Since there are three digits and that the average of any two should be the third digit or one of the digits, there are 9 possible averages i.e. digits(1,2,3 ...9). Lets see how many numbers each of these averages form to get us total required numbers.
1. 1 : 1+1 has 1 three digit number
2. 2 : 1 + 3 | 2 + 2 has (3! + 1) three digit number
3. 3 : 1 + 5 | 2 + 4 | 3 + 3 has (2*3! + 1) three digit number
4. 4 : 1 + 7 | 2 + 6 | 3 + 5 | 4 + 4 has (3*3! + 1) three digit number
5. 5 : 1 + 9 | 2 + 8 | 3 + 7 | 4 + 6 | 5 + 5 has (4*3! + 1) three digit number
6. 6 : 3 + 9 | 4 + 8 | 5 + 7 | 6 + 6 has (3*3! + 1) three digit number
7. 7 : 5 + 9 | 6 + 8 | 7 + 7 has (2*3! + 1) three digit number
8. 8 : 7 + 9 | 8 + 8 has (3! + 1) three digit number
9. 9 : 9 + 9 has 1 three digit number

Edit:
Additionally, 1, 2, 3 and 4 can be averages with other two being digits from 0 to 8.
1. 1 : 0 + 2 has 4 three digit number
2. 2 : 0 + 4 has 4 three digit number
3. 3 : 0 + 6 has 4 three digit number
4. 4 : 0 + 8 has 4 three digit number
Total = 4 * 4 = 16

G. Total = 1 + 3! + 1 + 2*3! + 1 + 3*3! + 1 + 4*3! + 1 + 3*3! + 1 + 2*3! + 1 + 3! + 1 + 1 + 16 = 16*6 + 9*1 + 16 = 121.

Answer D.
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Consider integer \(ABC\).
If A is equal to the average of B and C, we get:
\(A = \frac{B+C}{2}\)
\(2A = B+C\)
Implication:
Digits B and C must sum to twice the value of A.

A --> options for B and C
1 --> 0,2...............................................1,1
2 --> 0,4..........1,3................................2,2
3 --> 0,6..........1,5...2,4........................3,3
4 --> 0,8..........1,7...2,6...3,5................4,4
5 --> ...............1,9...2,8...3,7...4,6........5,5
6 --> ...............3,9...4,8...5,7................6,6
7 --> ...............5,9...6,8........................7,7
8 --> ...............7,9................................8,8
9 --> ....................................................9.9

Case 1: Digit combinations implied just to the right of the arrows
Number of cases in the column just to the right of the arrows = 4
Number of options for the hundreds digit = 2 (Cannot be 0)
Number of options for the tens digit = 2 (Either of the two remaining digits)
Number of options for the units digit = 1 (Only 1 digit left)
To combine these options, we multiply:
4*2*2*1 = 16

Case 2: Digit combinations implied in the middle of the chart
Number of cases in the middle of the chart = 16
Number of ways to arrange the 3 digits in each case = 3! = 6
To combine these options, we multiply:
16*6 = 96

Case 3: Digit combinations implied all the way to the right
Number of cases in the rightmost column = 9
Since the three digits in each case are identical, each case yields only one possible arrangement, for a total of 9 options.

Number of viable integers = Case 1 + Case 2 + Case 3 = 16 + 96 + 9 = 121


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