Consecutive primes whose quotient of their product and sum is itself a prime number.
$$ 2 \times 3 \times 5 = 30 $$
$$ 30/10 = 3 $$
$$ 3 \times 5 \times 7 = 105 $$
$$ 105/15 = 7 $$
Question: Amateur question, but is there a way to check for other consecutive primes which meet this criteria without having to write custom code like I see so often? What's the next sequence of primes like this? If I could learn a way do that kind of calculation for this and all future problems it would save me a lot of time.
Thanks!
Suppose $5\leq p<q<r$ are consecutive primes. Then $$ p+q+r<3r<pr<qr<pqr $$ so it must be that $$ p+q+r=pq. $$ By Bertrand's postulate, $r<2q$. It follows that $$ (p-1)q=p+r<q+2q=3q. $$ So it must be that $p+r=2q$ but then $p-1=2$ so that $p=3$. In other words, there are no solutions for $5\leq p<q<r$. The solutions you checked must be the only ones.
Generalizations: for four consecutive primes, you can either adapt the argument above or use this simple observation: suppose $p<q<r<s$ are consecutive primes. By inspection, $p=2$ doesn't work. So suppose $p>2$. But then, $p,q,r,s$ are all odd so that $p+q+r+s$ is even, which in turn implies $pqrs$ is even. This forces $p=2$. A contradiction.
You can now basically repeat this argument for the problem involving $n$ consecutive primes where $n$ is even. For $n$ is odd, pick a clever $N$ so that if $N\leq p_1<p_2<\ldots<p_n$ constitute a solution, then you can use Bertrand's postulate to force $$ p_1+p_2+\cdots+p_n=p_1p_2.\tag{*} $$ Using Bertrand's postulate once more, $$ Np_2\leq p_1p_2=p_1+p_2+\cdots +p_n<p_2+p_2+2p_2+\cdots 2^{n-2}p_2=2^{n-1}p_2. $$ This should narrow down the possibilities for $p_1$. I have seen that you can sharpen the bound $p_{n+1}<2p_n$ but I can't find a link for that anymore.
Final edit: any $N\geq 1+2^{n-2}$ works.