To solve the problem, we will denote the four natural numbers as \( A, B, C, \) and \( D \). We will use the information given about the averages to set up equations.
### Step-by-Step Solution:
1. **Set Up the Equations**:
- The average of \( A, B, C \) plus \( D \) gives us 29:
\[
\frac{A + B + C}{3} + D = 29
\]
Multiplying through by 3, we get:
\[
A + B + C + 3D = 87 \quad \text{(Equation 1)}
\]
- The average of \( B, C, D \) plus \( A \) gives us 23:
\[
\frac{B + C + D}{3} + A = 23
\]
Multiplying through by 3, we get:
\[
B + C + D + 3A = 69 \quad \text{(Equation 2)}
\]
- The average of \( C, D, A \) plus \( B \) gives us 21:
\[
\frac{C + D + A}{3} + B = 21
\]
Multiplying through by 3, we get:
\[
C + D + A + 3B = 63 \quad \text{(Equation 3)}
\]
- The average of \( D, A, B \) plus \( C \) gives us 17:
\[
\frac{D + A + B}{3} + C = 17
\]
Multiplying through by 3, we get:
\[
D + A + B + 3C = 51 \quad \text{(Equation 4)}
\]
2. **Add All Equations**:
- Adding all four equations:
\[
(A + B + C + 3D) + (B + C + D + 3A) + (C + D + A + 3B) + (D + A + B + 3C) = 87 + 69 + 63 + 51
\]
- This simplifies to:
\[
6(A + B + C + D) = 270
\]
- Dividing by 6:
\[
A + B + C + D = 45 \quad \text{(Equation 5)}
\]
3. **Find Individual Numbers**:
- Now, we will subtract Equation 1 from Equation 5:
\[
(A + B + C + D) - (A + B + C + 3D) = 45 - 87
\]
- This simplifies to:
\[
D - 2D = -42 \implies -D = -42 \implies D = 21
\]
4. **Substituting Back to Find Other Numbers**:
- Now, we can use \( D = 21 \) in Equation 1:
\[
A + B + C + 3(21) = 87
\]
\[
A + B + C + 63 = 87 \implies A + B + C = 24 \quad \text{(Equation 6)}
\]
- Now, substituting \( D = 21 \) into Equation 2:
\[
B + C + 21 + 3A = 69
\]
\[
B + C + 3A = 48 \quad \text{(Equation 7)}
\]
- From Equation 6, we can substitute \( C = 24 - A - B \) into Equation 7:
\[
B + (24 - A - B) + 21 + 3A = 69
\]
\[
24 - A + 21 + 3A = 69
\]
\[
24 + 21 + 2A = 69 \implies 2A = 24 \implies A = 12
\]
- Now substituting \( A = 12 \) back into Equation 6:
\[
12 + B + C = 24 \implies B + C = 12 \quad \text{(Equation 8)}
\]
- Now substituting \( A = 12 \) into Equation 7:
\[
B + C + 21 + 36 = 69 \implies B + C + 57 = 69 \implies B + C = 12
\]
- This confirms Equation 8.
5. **Finding \( B \) and \( C \)**:
- Since \( B + C = 12 \), we can choose values for \( B \) and \( C \) such that they are natural numbers. For example, if \( B = 6 \), then \( C = 6 \).
Thus, the four numbers are \( A = 12, B = 6, C = 6, D = 21 \).
### Final Answer:
One of the numbers is \( 21 \).
