Poirot1
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I can't see why the following is true:
Let $p_{1}<p_{2}<...<p_{k}$ be primes such that
$7(p_{1}-1)...(p_{k}-1)=3p_{1}...p_{k}$.
Since 7 divides the LHS, $p_{k}>or =7$
Let $p_{1}<p_{2}<...<p_{k}$ be primes such that
$7(p_{1}-1)...(p_{k}-1)=3p_{1}...p_{k}$.
Since 7 divides the LHS, $p_{k}>or =7$