Consider the system of equations `x_1+x_(2)^(2)+x_(3)^(3)+x_(4)^(4)+x_(5)^(5)=5` and `x_1+2x_2+3x_3+4x_4+5x_5=15` where `x_1,x_2,x_3,x_4,x_5` are positive real numbers. Then numbers of `(x_1,x_2,x_3,x_4,x_5)` is ___________.

Explanation

To solve the given system of equations, we have:

1. \( x_1 + x_2^2 + x_3^3 + x_4^4 + x_5^5 = 5 \) (Equation 1)
2. \( x_1 + 2x_2 + 3x_3 + 4x_4 + 5x_5 = 15 \) (Equation 2)

where \( x_1, x_2, x_3, x_4, x_5 \) are positive real numbers.

### Step 1: Analyze the equations

We know that all variables are positive real numbers. We can use the properties of inequalities to analyze the equations.

### Step 2: Apply the Cauchy-Schwarz Inequality

We can apply the Cauchy-Schwarz inequality to the second equation:

\[ (x_1 + x_2 + x_3 + x_4 + x_5)(1^2 + 2^2 + 3^2 + 4^2 + 5^2) \geq (x_1 + 2x_2 + 3x_3 + 4x_4 + 5x_5)^2 \]

Calculating \( 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55 \).

Thus, we have:

\[ (x_1 + x_2 + x_3 + x_4 + x_5)(55) \geq 15^2 \]

This simplifies to:

\[ x_1 + x_2 + x_3 + x_4 + x_5 \geq \frac{225}{55} = \frac{45}{11} \]

### Step 3: Relate the two equations

From Equation 1, we know:

\[ x_1 + x_2^2 + x_3^3 + x_4^4 + x_5^5 = 5 \]

Since \( x_i \) are positive, \( x_2^2, x_3^3, x_4^4, x_5^5 \) are all positive and contribute to the total of 5.

### Step 4: Use the AM-GM Inequality

We can also apply the AM-GM inequality to Equation 1:

\[ \frac{x_1 + x_2^2 + x_3^3 + x_4^4 + x_5^5}{5} \geq \sqrt[5]{x_1 \cdot x_2^2 \cdot x_3^3 \cdot x_4^4 \cdot x_5^5} \]

This implies:

\[ \frac{5}{5} \geq \sqrt[5]{x_1 \cdot x_2^2 \cdot x_3^3 \cdot x_4^4 \cdot x_5^5} \]

Thus:

\[ 1 \geq \sqrt[5]{x_1 \cdot x_2^2 \cdot x_3^3 \cdot x_4^4 \cdot x_5^5} \]

Raising both sides to the power of 5 gives:

\[ 1 \geq x_1 \cdot x_2^2 \cdot x_3^3 \cdot x_4^4 \cdot x_5^5 \]

### Step 5: Finding the values

To find the number of solutions, we need to check if there are specific values of \( x_1, x_2, x_3, x_4, x_5 \) that satisfy both equations.

Assume \( x_1 = x_2 = x_3 = x_4 = x_5 = k \). Then:

1. \( k + k^2 + k^3 + k^4 + k^5 = 5 \)
2. \( k + 2k + 3k + 4k + 5k = 15k = 15 \Rightarrow k = 1 \)

Substituting \( k = 1 \) into the first equation:

\[ 1 + 1^2 + 1^3 + 1^4 + 1^5 = 1 + 1 + 1 + 1 + 1 = 5 \]

This satisfies both equations.

### Conclusion

The only solution is \( (1, 1, 1, 1, 1) \). Therefore, the number of solutions \( (x_1, x_2, x_3, x_4, x_5) \) is:

**1**