MHB Is the Dividing Problem Involving Primes Solvable?

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The discussion centers on the equation involving primes, specifically 7(p1-1)...(pk-1) = 3p1...pk. It is established that if pk is the largest prime, then pk must be greater than or equal to 7 due to the division by 7 on the left-hand side. The argument further concludes that if p1 is not equal to 2, the terms would yield an even product on one side and an odd product on the other, leading to the necessity of p1 being 2. Consequently, this implies k must equal 2, resulting in the conclusion that p2 must be 7. The discussion ultimately explores the conditions under which the dividing problem involving primes can be solved.
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, $p_{k}>or =7$
 
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Re: dividing problem

Poirot said:
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, $p_{k}>or =7$

If $p_{1} \ne 2$ then the term $(p_{1}-1)\ ...\ (p_{k}-1)$ is even and the term $p_{1}\ ...\ p_{k}$ is odd ... that's impossible so that it must be $p_{1}=2$. In this case the term $(p_{1}-1)\ ...\ (p_{k}-1)$ contains as factor $2^{k-1}$ and the term $p_{1}\ ...\ p_{k}$ contains as factor 2, so that it must be k=2. In this case it is... $\displaystyle 7\ (p_{2}-1)= 3\ 2\ p_{2} \implies p_{2}=7$ (2)Kind regards$\chi$ $\sigma$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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