I have been able to find the answer via exhaustion. I defined $a\le b\le c\le d\le e\le f$ as the primes and solved an inequality for each variable. The answer is $2,2,3,3,7,67$. However, the number of cases to solve got quite large and I was wondering if there is any neat solution to this problem.
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6$\begingroup$ $201=3\times67$, and the product is divisible by $201$, so $3$ and $67$ are two of the primes $\endgroup$– J. W. TannerNov 12, 2020 at 0:31
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$\begingroup$ If the product is $201$ times larger than the sum, then the product is $202$ times as large as the sum, so the product is divisible by $202=2\times101$, so one of the primes is $101$. So either your solution is wrong, or the problem statement is wrong. $\endgroup$– Gerry MyersonNov 12, 2020 at 2:06
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$\begingroup$ I guess this isn't a solution but its something very interesting to consider. Writing your problem as an equation: $$abcdef=201(a+b+c+d+e+f)$$ by AM-GM, the RHS must satisfy the following: $$\frac{a+b+c+d+e+f}{6}\geq(abcdef)^\frac{1}{6}$$ $$201(a+b+c+d+e+f)\geq 201\times6\times(abcdef)^\frac{1}{6}$$ $$abcdef=201(a+b+c+d+e+f)\geq 201\times6\times(abcdef)^\frac{1}{6}$$ $$abcdef\geq 3\times67\times2\times3\times(6\times201)^\frac{1}{5}$$ The final term works out to ~4.13. That is to say that your product has a lower bound, intrestingly this is very close to your answer. $\endgroup$– personNov 12, 2020 at 2:21
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$\begingroup$ @person, $201$ times larger is $202$ times as large. $\endgroup$– Gerry MyersonNov 12, 2020 at 12:07
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1 Answer
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Start with the observation of J. W. Tanner that two of the primes must be $3,67$. Note that if all of the primes are odd, their product is odd, but their sum is even, so we can conclude that one of the primes must be $2$.
Restate the problem as $2\cdot 3\cdot 67\cdot p_1p_2p_3=201(p_1+p_2+p_3+72)$ from which $2\cdot p_1p_2p_3=(p_1+p_2+p_3+72)$. Once again, if all unidentified primes are odd then LHS is even but RHS is odd, so there is another factor of $2$ present. This yields $4\cdot p_1p_2=(p_1+p_2+74)$. We can quickly convince ourselves that $p_1\ne p_2$ and $p_1\equiv p_2 \bmod 4$.
The smallest pair of primes that are $\equiv 1 \bmod 4$ are $5,13$, but $4\cdot 5\cdot 13>5+13+74$ and looking at larger primes only makes the inequality greater. The smallest pair of primes that are $\equiv 3 \bmod 4$ are $3,7$, and this provides a solution. The next larger pair of primes that are $\equiv 3 \bmod 4$ are $7,11$, and also in this case, $4\cdot 7\cdot 11>7+11+74$ and looking at larger primes only makes the inequality greater. So the remaining primes are $3,7$.
The complete list is as OP found: $2,2,3,3,7,67$.
answered Nov 12, 2020 at 3:47
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3$\begingroup$ From $4p_1p_2=p_1+p_2+74$, you get $(4p_1-1)(4p_2-1)=297$ $\endgroup$– Empy2Nov 12, 2020 at 4:12
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1$\begingroup$ But $201$ times larger is $202$ times as large. $\endgroup$ Nov 12, 2020 at 12:08
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$\begingroup$ $(3,11)$ is smaller than $(7,11)$ though. But since your argument is that the difference increases with $p_1$ and $p_2$, you shouldn't need to consider other cases than $(3,7)$ at all. $\endgroup$– MiltenNov 12, 2020 at 14:37