Find all prime numbers $p_{1}$, $p_{2}$, $p_{3}$ $\ldots$ $p_{n}$ such that $\prod_{i=1}^{n} p_{i} = 10 \sum_{i=1}^{n} p_{i}$

Given a quadrilateral $ABCD$ with $\angle B=\angle D=90^{\circ}$. Point $M$ is chosen on segment $AB$ so that $AD=AM$. Rays $DM$ and $CB$ intersect at point $N$. Points $H$ and $K$ are feet of perpendiculars from points $D$ and $C$ to lines $AC$ and $AN$ respectively. Prove that $\angle MHN=\angle MCK$.

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TopNewestQ1)A bit of a sleepy proof (I'm gonna sleep soon). Tell me my errors and I'll rectify them.Firstly, note that the LHS must be a multiple of 10, so 2 of the primes (we'll call them $p_1$ and $p_2$ are 2 and 5. We now have

$p_1 p_2 \ldots p_n = 10(p_1+p_2+\ldots+p_n)$

$p_3 p_4 \ldots p_n = 7+p_3+p_4+\ldots+p_n$

WLOG $p_3 \leq p_4 \leq \ldots \leq p_n$

We will now prove that there is only a maximum of 4 terms in this sequence. Assume that there are $k$ terms for $k \geq 5$.

$p_3 p_4 \ldots p_k = p_3 + p_4 + \ldots + p_k+7$

We have $p_3+p_4+\ldots+p_k +7 \leq (k-2) p_k + 7$ and $p_3 p_4 \ldots p_k \leq p_k^{k-2}$. We can prove via induction (or other methods) that $(k-2)p_k+7 < p_k^{k-2}$ for $p_k \geq 3, k \geq 5$. Thus, no solutions exist when there are 5 or more terms.

We now have 3 cases:

Case 1:4 terms in the sequence$p_1 p_2 p_3 p_4 = 10(p_1+p_2+p_3+p_4)$

$p_3 p_4 = 7 + p_3 + p_4$

$(p_3 - 1)(p_4 - 1) = 8$

We must have $p_3 - 1 =2$ and $p_4-1=4$ so $p_3=3$ and $p_4 =5$. Thus, we have solution $2, 3, 5, 5$.

Case 2:3 terms in the sequence$p_1 p_2 p_3 = 10(p_1+p_2+p_3)$

$p_3 = 7 + p_3$

No solutions here.

Case 3:2 terms in the sequence$p_1 p_2 = 10(p_1+p_2)$

$1 = 7$

No solutions here.

Therefore, only $2, 3, 5, 5$ is the solution that satisfies.

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Just for clarification, in Q1, can two primes be the same, i.e. is it possible to have 5 and 5 as two of the primes?

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Yes

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