Find all primes such that $\prod_{i=1}^{n} p_i=10\sum_{i=1}^{n}p_i$ Find all primes $p$ (it's not necessary to this primes to be different) such that $$\prod_{i=1}^{n} p_i=10\sum_{i=1}^{n}p_i$$
I’ve realized that $$10\mid \prod_{i=1}^{n} p_i\implies \prod_{i=1}^{n} p_i =2\cdot5...$$
But i don’t know what is next.
 A: Using your observation, the equation can be reduced to (here I replaced $n-2$ with $n$):
$$\prod_{i=1}^n p_i = 7+\sum_{i=1}^np_i$$
It helps to first establish upper and lower bounds for the expressions. The product grows much faster than the sum, so we will find a lower bound for the product and an upper bound for the sum.
Let $p$ be the largest prime among $p_i$. Then we have
$$\prod_{i=1}^n p_i \ge 2^{n-1}p, \quad \sum_{i=1}^np_i \le np$$
therefore there are no solutions if $2^{n-1}p > 7+np$.
Obviously $p\ne 2$ and $n\ne 1$. Since $p\ge 3$, we can check that $n < 4$. This helps reduce the number of cases to check.
For $n=2$, we have the equation $pq = 7 + p + q$, or $(p-1)(q-1)=8$. By consider the factors of $8$, we must have $p=5, q=3$ (and not $p=9, q=2$).
For $n=3$, by a parity argument we see that there is exactly one $2$ in $p_i$.  Hence we have the equation
$$2pq = 7+2 + p + q \leadsto4pq-2p-2q+1 = 18+1 \leadsto(2p-1)(2q-1) = 19$$
but by considering the factors of $19$, we see that a prime solution to $(p,q)$ is impossible.
Therefore the only set of primes satisfying our original equation is $(2,3,5,5)$, with
$$2\times 3 \times 5 \times 5 = 10 (2+3+5+5)$$
