Let $n$ be a two-digit natural number such that the remainder of $n ^ { 3 }$ divided by 66 is $n$. We are to find the number of such $n$ 's and to find the prime numbers among them.
From the conditions we have
$$n ^ { 3 } = \mathbf { A B } p + n \quad ( 0 < n \leqq \mathbf { C D } ) ,$$
where $p$ is the integer quotient of $n ^ { 3 }$ divided by 66 . This can be transformed into
$$n ( n - 1 ) ( n + 1 ) = \mathrm { AB } p$$
Since either $n - 1$ or $n$ has to be a multiple of $\mathbf { E }$ and either $n - 1 , n$ or $n + 1$ has to be a multiple of $\mathbf { F }$, and furthermore $\mathbf { E }$ and $\mathbf { F }$ are mutually prime, we know that $n ( n - 1 ) ( n + 1 )$ is a multiple of $\mathbf { G }$. (Write the answers in the order $1 < \square < \mathbf { E } < \mathbf { F } < \mathbf { G }$.) Hence one of $n - 1 , n$ and $n + 1$ must be a multiple of $\mathbf{HI}$.
So, since $n \leqq \mathrm { CD }$, the number of $n$ 's where $n - 1$ is a multiple of $\mathbf{HI}$ is $\mathbf{J}$, where $n$ is a multiple of $\mathbf{HI}$ is $\mathbf { K }$, and where $n + 1$ is a multiple of $\mathbf{HI}$ is $\mathbf { L }$.
Thus, the number of $n$ 's is $\mathbf { M N }$ and the prime numbers among them are $\mathbf { O P } , \mathbf { Q R }$, $\mathbf{ST}$, in ascending order.