# The product of which 6 primes is 201-times larger than their sum?

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.

• $201=3\times67$, and the product is divisible by $201$, so $3$ and $67$ are two of the primes Nov 12, 2020 at 0:31
• 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. Nov 12, 2020 at 2:06
• 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. Nov 12, 2020 at 2:21
• @person, $201$ times larger is $202$ times as large. Nov 12, 2020 at 12:07

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: $$2,2,3,3,7,67$$.
• From $4p_1p_2=p_1+p_2+74$, you get $(4p_1-1)(4p_2-1)=297$ Nov 12, 2020 at 4:12
• But $201$ times larger is $202$ times as large. Nov 12, 2020 at 12:08
• $(3,11)$ 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. Nov 12, 2020 at 14:37