Problem 843

Solution:
Because the product is approximately $64 \cdot 10^{30}$, we know the primes are all around $200000$. Say they are $200000 + x_{i}$ for $i = 1, \ldots, 6$.
By expanding $\prod_{i=1}^{6}\left(200000 + x_{i}\right)$ as a polynomial in $200000$, we see that
$$31231 \cdot 10^{25} = 200000^{5}\left(x_{1} + \cdots + x_{6}\right)$$
plus the carry from the other terms. Note that $31231 = 975 \cdot 32 + 31$, so $x_{1} + \cdots + x_{6} \leq 975$.
Thus,
$$16\left(x_{1} x_{2} + x_{1} x_{3} + \cdots + x_{5} x_{6}\right) \leq 16 \cdot \frac{5}{12}\left(x_{1} + \cdots + x_{6}\right)^{2} < \frac{20}{3} \cdot 1000^{2} < 67 \cdot 10^{5}$$
so the carry term from $200000^{4}\left(x_{1} x_{2} + \cdots + x_{5} x_{6}\right)$ is at most $67 \cdot 10^{25}$. The other terms have negligible carry, so it is pretty clear $x_{1} + \cdots + x_{6} > 972$, otherwise the carry term would have to be at least
$$31231 \cdot 10^{25} - 200000^{5}(972) = 127 \cdot 10^{25}$$
It follows that $x_{1} + \cdots + x_{6}$ lies in $[973, 975]$, so the sum of the primes, $6 \cdot 200000 + (x_{1} + \cdots + x_{6})$, lies in $[1200973, 1200975]$.
As these primes are all greater than $2$, they are all odd, so their sum is even. Thus it must be $1200974$.