Find all prime numbers $p_1,…,p_n$ (not necessarily different) such that :
$$ \prod_{i=1}^n p_i=10 \sum_{i=1}^n p_i$$
1. We start with the given equation:
   \[
   \prod_{i=1}^n p_i = 10 \sum_{i=1}^n p_i
   \]
   Since \(10\) divides the right-hand side (RHS), we must have at least one factor of \(2\) and one factor of \(5\) among the \(p_i\). Without loss of generality, let \(p_1 = 2\) and \(p_2 = 5\).

2. Substituting \(p_1 = 2\) and \(p_2 = 5\) into the equation, we get:
   \[
   2 \cdot 5 \cdot \prod_{i=3}^n p_i = 10 \left(2 + 5 + \sum_{i=3}^n p_i\right)
   \]
   Simplifying, we have:
   \[
   10 \prod_{i=3}^n p_i = 10 \left(7 + \sum_{i=3}^n p_i\right)
   \]
   Dividing both sides by \(10\), we obtain:
   \[
   \prod_{i=3}^n p_i = 7 + \sum_{i=3}^n p_i
   \]

3. We now consider the case \(n = 2\):
   \[
   \prod_{i=1}^2 p_i = 10 \sum_{i=1}^2 p_i
   \]
   This simplifies to:
   \[
   2 \cdot 5 = 10 (2 + 5)
   \]
   \[
   10 = 70
   \]
   This is clearly false, so \(n \neq 2\).

4. Next, we try \(n = 4\):
   \[
   \prod_{i=3}^4 p_i = 7 + \sum_{i=3}^4 p_i
   \]
   Let \(q_1 = p_3\) and \(q_2 = p_4\). Then:
   \[
   q_1 q_2 = 7 + q_1 + q_2
   \]
   Rearranging, we get:
   \[
   q_1 q_2 - q_1 - q_2 = 7
   \]
   Adding \(1\) to both sides:
   \[
   (q_1 - 1)(q_2 - 1) = 8
   \]
   The factor pairs of \(8\) are \((1, 8)\), \((2, 4)\), \((4, 2)\), and \((8, 1)\). Since \(q_1\) and \(q_2\) are primes, the only valid pairs are \((3, 5)\) and \((5, 3)\).

5. Therefore, the possible values for \(p_3\) and \(p_4\) are \(3\) and \(5\). Thus, the solution is:
   \[
   (p_1, p_2, p_3, p_4) = (2, 3, 5, 5)
   \]

6. To verify, we substitute back into the original equation:
   \[
   2 \cdot 3 \cdot 5 \cdot 5 = 10 (2 + 3 + 5 + 5)
   \]
   \[
   150 = 10 \cdot 15
   \]
   \[
   150 = 150
   \]
   This confirms that the solution is correct.

The final answer is \( \boxed{ (2, 3, 5, 5) } \)