1957 AHSME Problem 38

From a two-digit number N we subtract the number with the digits reversed and find that the result is a positive perfect cube. Then:

\begin{array}{l} \textbf{(A)}\ {N}\text{ cannot end in 5}\qquad\\ \textbf{(B)}\ {N}\text{ can end in any digit other than 5}\qquad \\ \textbf{(C)}\ {N}\text{ does not exist}\qquad\\ \textbf{(D)}\ \text{there are exactly 7 values for }{N}\qquad\\ \textbf{(E)}\ \text{there are exactly 10 values for }{N} \end{array}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.