, 2022

Solution:

Let $B = b_{1} b_{2} \cdots b_{7} - 128$. Since
$$a_{1}^{b_{1} b_{2} \cdots b_{7}} \equiv a_{2}^{2 b_{2} b_{3} \cdots b_{7}} \equiv a_{3}^{4 b_{3} b_{4} \cdots b_{7}} \equiv \cdots \equiv a_{1}^{128} \quad(\bmod 167)$$
we find that $a_{1}^{B} \equiv 1 (\bmod 167)$. Similarly, $a_{i}^{B} \equiv 1 (\bmod 167)$ for all $i$. Since 167 is a prime and $167-1=2 \cdot 83$, we know that the order of each individual $a_{i}$ (since $a_{i} \neq 1$) must be either 2 or a multiple of 83. If $B$ is not a multiple of 83, then it follows that all the $a_{i}$ must be $-1$, which implies that all the $b_{i}$ must be even, meaning that the minimum possible value of $b_{1} b_{2} \cdots b_{7}\left(b_{1}+b_{2}+\cdots+b_{7}\right)$ is $2^{7} \cdot 14 > 1000$.

On the other hand, if $B$ is a multiple of 83, then the smallest possible values for $b_{1} b_{2} \cdots b_{7}$ are 45 and 128. If $b_{1} b_{2} \cdots b_{7}=45$, then the smallest possible value for $b_{1}+b_{2}+\cdots+b_{7}$ is $5+3+3+1+1+1+1=15$, so the minimum possible value for $b_{1} b_{2} \cdots b_{7}\left(b_{1}+b_{2}+\cdots+b_{7}\right)$ is $45 \cdot 15=675$. This can be achieved by letting $g$ be an element of order 83 and setting $a_{1}=g$, $a_{2}=g^{1 / 2}$, $a_{3}=g^{1 / 4}$, $a_{4}=g^{1 / 8}$, $a_{5}=g^{1 / 16}$, $a_{6}=g^{3 / 32}$, $a_{7}=g^{9 / 64}$ (all exponents are taken mod 83).

If $b_{1} b_{2} \cdots b_{7} \geq 128$, then by the AM-GM inequality we have
$$b_{1} b_{2} \cdots b_{7}\left(b_{1}+b_{2}+\cdots+b_{7}\right) \geq 7\left(b_{1} b_{2} \cdots b_{7}\right)^{8 / 7} \geq 7 \cdot 2^{8} > 1000$$
Therefore 675 is optimal.