DylanN wrote: Let $k$ be a positive integer. Consider $k$ not necessarily distinct prime numbers such that their product is ten times their sum. What are these primes and what is the value of $k$? Product of the $k$ numbers is $n=p_1^{a_1}p_2^{a_2}.....p_k^{a_k}$ Given $n=p_1^{a_1}p_2^{a_2}.....p_k^{a_k}=10(p_1^{a_1}+...p_k^{a_k}$. So one of the prime is $2$ and other is $5$. So $2\times5p_1^{a_3}\times... \times p_k^{a_k}=10(p_1^{a_1}+...p_k^{a_k})$ So it turns out that the product of $8$ numbers is equal to the sum of $10$ numbers, $8$ of which are similar. Now just case check to see that how many of them are $1$ and other numbers etc.

We suppose WLOG that $p_1 \le p_2 \le ... p_k$ $p_1p_2...p_k=10(p_1+p_2+...p_k)$ thus $p_1=2 \quad p_l=5 \qquad l\le k$ $\implies p_2p_3..p_{l-1}p_{l+1}..p_k=7+p_2+p_3+..p_{l-1}+p_{l+1}+..p_k$ $p_2p_3..p_{l-1}p_{l+1}..p_k\le 7+(k-2)p_k \qquad p_k\ge 5$ $\implies p_2p_3..p_{l-1}p_{l+1}..p_{k-1}\le k-1$ $\implies 2^{k-3} < k-1$ Therefore $k\in [{3;4}]$ if $k=3$ then $p_m=7+p_m\qquad m\le k$ wich is impossible So $k=4$ Then $p_2p_{3/4}=7+p_2+p_{3/4}$ $\implies (p_2-1)(p_{3/4}-1)=8$ Thus $p_1=2,p_2=3,p_3=p_4=5$

Let the primes be $p_1,p_2,...,p_k$. We know that: $$10 \sum_{i=1}^kp_i = \prod_{i=1}^kp_i$$Therefore, two of the primes must be $2$ and $5$. Assume that $p_1=2$ and $p_2 = 5$. It is not hard to see that: $$7 + \sum_{i=3}^kp_i = \prod_{i=3}^kp_i$$Now, let $\sum_{i=3}^kp_i = x$ and $\prod_{i=3}^kp_i = y$. We know that by AM-GM $\frac{x}{k-2} \ge \sqrt[k-2]{y} \implies \left( \frac{x}{k-2} \right)^{k-2} \ge y$. Therefore, we have that: $$7 + x= y \implies \left( \frac{x}{k-2} \right)^{k-2} \le 7 + x$$It’s not hard to see that this implies $k = 4$. If $k=4$, then we have that $7 + p_3 + p_4 = p_3p_4 \implies (p_3-1)(p_4-1) = 8 \implies (p_3,p_4) = (3,5),(5,3)$. Therefore, the only possible primes are permutations of $(2,3,5,5)$.

How did you get from Beyourself wrote: ... $p_2p_3..p_{l-1}p_{l+1}..p_k\le 7+(k-2)p_k \qquad p_k\ge 5$ ... to Beyourself wrote: ... $\implies p_2p_3..p_{l-1}p_{l+1}..p_{k-1}\le k-1$ ...

AopsUser101 wrote: ... Therefore, we have that: $$7 + x= y \implies \left( \frac{x}{k-2} \right)^{k-2} \le 7 + x$$... Shouldn't it be $$7 + x= y \implies \left( \frac{x}{k-2} \right)^{k-2} \ge 7 + x$$