Let a,b,c,d and e be positive real numbers such that `a+b+c+d+e=15` and `ab^2c^3d^4e^5=(120)^3xx50`. Then the value of `a^2+b^2+c^2+d^2+e^2` is ___________.

Explanation

To solve the problem, we will use the Arithmetic Mean-Geometric Mean (AM-GM) inequality and the given conditions.

### Step-by-Step Solution:

1. **Given Information:**
We have five positive real numbers \(a, b, c, d, e\) such that:
\[ a + b + c + d + e = 15 \]
and
\[ ab^2c^3d^4e^5 = (120)^3 \times 50. \]

2. **Rewriting the Product:**
First, we simplify the right-hand side:
\[ (120)^3 = (2^3 \times 3^3 \times 5)^3 = 2^9 \times 3^9 \times 5^3, \]
and
\[ 50 = 2 \times 5^2. \]
Therefore,
\[ (120)^3 \times 50 = 2^9 \times 3^9 \times 5^3 \times (2 \times 5^2) = 2^{10} \times 3^9 \times 5^5. \]

3. **Using AM-GM Inequality:**
By the AM-GM inequality, we have:
\[ \frac{a + \frac{b}{2} + \frac{b}{2} + \frac{c}{3} + \frac{c}{3} + \frac{c}{3} + \frac{d}{4} + \frac{d}{4} + \frac{d}{4} + \frac{d}{4} + \frac{e}{5} + \frac{e}{5} + \frac{e}{5} + \frac{e}{5} + \frac{e}{5}}{15} \geq \sqrt[15]{ab^2c^3d^4e^5}. \]
Here, we have \(15\) terms in total.

4. **Calculating the Left Side:**
The left side simplifies to:
\[ \frac{a + b + c + d + e}{15} = \frac{15}{15} = 1. \]

5. **Calculating the Right Side:**
The right side is:
\[ \sqrt[15]{ab^2c^3d^4e^5} = \sqrt[15]{2^{10} \times 3^9 \times 5^5} = 2^{\frac{10}{15}} \times 3^{\frac{9}{15}} \times 5^{\frac{5}{15}} = 2^{\frac{2}{3}} \times 3^{\frac{3}{5}} \times 5^{\frac{1}{3}}. \]

6. **Setting Up the Equality:**
For the AM-GM inequality to hold with equality, all the terms must be equal:
\[ a = \frac{b}{2} = \frac{c}{3} = \frac{d}{4} = \frac{e}{5} = \lambda. \]
Thus, we can express \(b, c, d, e\) in terms of \(a\):
\[ b = 2\lambda, \quad c = 3\lambda, \quad d = 4\lambda, \quad e = 5\lambda. \]

7. **Substituting Back:**
Substituting these into the sum:
\[ a + 2\lambda + 3\lambda + 4\lambda + 5\lambda = 15 \implies a + 14\lambda = 15. \]
Since \(a = \lambda\), we have:
\[ \lambda + 14\lambda = 15 \implies 15\lambda = 15 \implies \lambda = 1. \]
Therefore, we find:
\[ a = 1, \quad b = 2, \quad c = 3, \quad d = 4, \quad e = 5. \]

8. **Calculating \(a^2 + b^2 + c^2 + d^2 + e^2\):**
Now we calculate:
\[ a^2 + b^2 + c^2 + d^2 + e^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55. \]

### Final Answer:
The value of \(a^2 + b^2 + c^2 + d^2 + e^2\) is **55**.

---