If the product of `n` positive numbers is `n^n` , then their sum is `a` positive integer b. divisible by `n` equal to `n+1//n` never less than `n^2` ``

Options

A. a positive integer
B. divisible by n
C. equal to `n+l//n`
D. never less than `n^2`

Explanation

To solve the problem, we need to analyze the given conditions using the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
We have `n` positive numbers whose product is equal to `n^n`. Let's denote these numbers as \( a_1, a_2, \ldots, a_n \). Thus, we have:
\[ a_1 \times a_2 \times \ldots \times a_n = n^n \]

2. **Applying the AM-GM Inequality:**
According to the AM-GM inequality, the arithmetic mean of `n` positive numbers is greater than or equal to the geometric mean of those numbers. This can be expressed as:
\[ \frac{a_1 + a_2 + \ldots + a_n}{n} \geq \sqrt[n]{a_1 \times a_2 \times \ldots \times a_n} \]
Substituting the product we have:
\[ \frac{a_1 + a_2 + \ldots + a_n}{n} \geq \sqrt[n]{n^n} \]

3. **Calculating the Geometric Mean:**
The geometric mean simplifies as follows:
\[ \sqrt[n]{n^n} = n \]
Therefore, we can rewrite the inequality:
\[ \frac{a_1 + a_2 + \ldots + a_n}{n} \geq n \]

4. **Multiplying Both Sides by `n`:**
To eliminate the fraction, we multiply both sides by `n`:
\[ a_1 + a_2 + \ldots + a_n \geq n^2 \]

5. **Conclusion:**
From the inequality derived, we conclude that the sum of the `n` positive numbers is never less than \( n^2 \). Therefore, the correct option is:
\[ \text{d) never less than } n^2 \]