Let $n _ { 1 } , n _ { 2 } , \cdots , n _ { 51 }$ be distinct natural numbers each of which has exactly 2023 positive integer factors. For instance, $2 ^ { 2022 }$ has exactly 2023 positive integer factors $1,2,2 ^ { 2 } , \cdots , 2 ^ { 2021 } , 2 ^ { 2022 }$. Assume that no prime larger than 11 divides any of the $n _ { i }$'s. Show that there must be some perfect cube among the $n _ { i }$'s. You may use the fact that $2023 = 7 \times 17 \times 17$.